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Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 7 Application of Definite Integration Miscellaneous Exercise 7 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 7 Application of Definite Integration Miscellaneous Exercise 7

(I) Choose the correct alternatives:

Question 1.
Area of the region bounded by the curve x² = y, the X-axis and the lines x = 1 and x = 3 is
(a) \(\frac{26}{3}\) sq units
(b) \(\frac{3}{26}\) sq units
(c) 26 sq units
(d) 3 sq units
Answer:
(a) \(\frac{26}{3}\) sq units

Question 2.
The area of the region bounded by y² = 4x, the X-axis and the lines x = 1 and x = 4 is
(a) 28 sq units
(b) 3 sq unit
(c) \(\frac{28}{3}\) sq units
(d) \(\frac{3}{28}\) sq units
Answer:
(c) \(\frac{28}{3}\) sq units

Question 3.
Area of the region bounded by x² = 16y, y = 1 and y = 4 and the Y-axis, lying in the first quadrant is
(a) 63 sq units
(b) \(\frac{3}{56}\) sq units
(c) \(\frac{56}{3}\) sq units
(d) \(\frac{63}{7}\) sq units
Answer:
(c) \(\frac{56}{3}\) sq units

Question 4.
Area of the region bounded by y = x⁴, x = 1, x = 5 and the X-axis is
(a) \(\frac{3142}{5}\) sq units
(b) \(\frac{3124}{5}\) sq units
(c) \(\frac{3142}{3}\) sq units
(d) \(\frac{3124}{3}\) sq units
Answer:
(b) \(\frac{3124}{5}\) sq units

Question 5.
Using definite integration area of circle x² + y² = 25 is
(a) 5π sq units
(b) 4π sq units
(c) 25π sq units
(d) 25 sq units
Answer:
(c) 25π sq units

(II) Fill in the blanks:

Question 1.
Area of the region bounded by y = x⁴, x = 1, x = 5 and the X-axis is _________
Answer:
\(\frac{3124}{5}\) sq units

Question 2.
Using definite integration area of the circle x² + y² = 49 is ___________
Answer:
49π sq units

Question 3.
Area of the region bounded by x² = 16y, y = 1, y = 4 and the Y-axis lying in the first quadrant is _________
Answer:
\(\frac{56}{3}\) sq units

Question 4.
The area of the region bounded by the curve x² = y, the X-axis and the lines x = 3 and x = 9 is _________
Answer:
234 sq units

Question 5.
The area of the region bounded by y² = 4x, the X-axis and the lines x = 1 and x = 4 is __________
Answer:
\(\frac{28}{3}\) sq units

(III) State whether each of the following is True or False.

Question 1.
The area bounded by the curve x = g(y), Y-axis and bounded between the lines y = c and y = d is given by \(\int_{c}^{d} x d y=\int_{y=c}^{y=d} g(y) d y\)
Answer:
True

Question 2.
The area bounded by two curves y = f(x), y = g(x) and X-axis is \(\left|\int_{a}^{b} f(x) d x-\int_{b}^{a} g(x) d x\right|\)
Answer:
False

Question 3.
The area bounded by the curve y = f(x), X-axis and lines x = a and x = b is \(\left|\int_{a}^{b} f(x) d x\right|\)
Answer:
True

Question 4.
If the curve, under consideration, is below the X-axis, then the area bounded by curve, X-axis, and lines x = a, x = b is positive.
Answer:
False

Question 5.
The area of the portion lying above the X-axis is positive.
Answer:
True

(IV) Solve the following:

Question 1.
Find the area of the region bounded by the curve xy = c², the X-axis, and the lines x = c, x = 2c.
Solution:

= c² log(\(\frac{2 c}{c}\))
= c² . log 2 sq units.

Question 2.
Find the area between the parabolas y² = 7x and x² = 7y.
Solution:

For finding the points of intersection of the two parabolas,
we equate the values of y² from their equations.
From the equation x² = 7y, y² = \(\frac{x^{4}}{49}\)
∴ \(\frac{x^{4}}{49}\) = 7x
∴ x⁴ = 343x
∴ x⁴ – 343x = 0
∴ x(x³ – 343) = 0
∴ x = 0 or x³ = 343, i.e. x = 7
When x = 0, y = 0
When x = 7, 7y = 49
∴ y = 7
∴ the points of intersection are O(0, 0) and A(7, 7)
Required area = area of the region OBACO
= (area of the region ODACO) – (area of the region ODABO)
Now, area of the region ODACO = area under the parabola y² = 7x
i.e. y = √7 √x

Area of the region ODABO = Area under the parabola
x² = 7y
i.e. y = \(\frac{x^{2}}{7}\)

∴ required area = \(\frac{98}{3}-\frac{49}{3}=\frac{49}{3}\) sq units.

Question 3.
Find the area of the region bounded by the curve y = x² and the line y = 10.
Solution:

By the symmetry of the parabola,
the required area is twice the area of the region OABCO
Now, the area of the region OABCO

Question 4.
Find the area of the ellipse \(\frac{x^{2}}{16}+\frac{y^{2}}{9}\) = 1.
Solution:
By the symmetry of the ellipse, the required area of the ellipse is 4 times the area of the region OPQO.
For the region OPQO, the limits of integration are x = 0 and x = 4.

Question 5.
Find the area of the region bounded by y = x², the X-axis and x = 1, x = 4.
Solution:
Required area = \(\int_{1}^{4} y d x\), where y = x²
= \(\int_{1}^{4} x^{2} d x\)
= \(\left[\frac{x^{3}}{3}\right]_{1}^{4}=\frac{4^{3}}{3}-\frac{1}{3}=\frac{64-1}{3}\)
= 21 sq units.

Question 6.
Find the area of the region bounded by the curve x² = 25y, y = 1, y = 4, and the Y-axis.
Solution:

Question 7.
Find the area of the region bounded by the parabola y² = 25x and the line x = 5.
Solution:

Given the equation of the parabola is y² = 25x
∴ y = 5√x …… [∵ IIn first quadrant, y > 0]
Required area = area of the region OQRPO
= 2(area of the region ORPO)

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 7 Application of Definite Integration Ex 7.1 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 7 Application of Definite Integration Ex 7.1

Question 1.
Find the area of the region bounded by the following curves, the X-axis, and the given lines:
(i) y = x⁴, x = 1, x = 5
Solution:

(ii) y = \(\sqrt{6 x+4}\), x = 0, x = 2
Solution:

(iii) \(\sqrt{16-x^{2}}\), x = 0, x = 4
Solution:

(iv) 2y = 5x + 7, x = 2, x = 8
Solution:

(v) 2y + x = 8, x = 2, x = 4
Solution:

(vi) y = x² + 1, x = 0, x = 3
Solution:

(vii) y = 2 – x², x = -1, x = 1
Solution:

Question 2.
Find the area of the region bounded by the parabola y² = 4x and the line x = 3.
Solution:

Required area = area of the region OABO
= 2(area of the region OACO)

Question 3.
Find the area of the circle x² + y² = 25.
Solution:

By the symmetry of the circle, its area is equal to 4 times the area of the region OABO.
Clearly, for this region, the limits of integration are 0 and 5.
From the equation of the circle, y² = 25 – x².
In the first quadrant y > 0
∴ y = \(\sqrt{25-x^{2}}\)
∴ area of the circle = 4(area of region OABO)

Question 4.
Find the area of the ellipse \(\frac{x^{2}}{4}+\frac{y^{2}}{25}\) = 1
Solution:

By the symmetry of the ellipse, its area is equal to 4 times the area of the region OABO.
Clearly, for this region, the limits of integration are 0 and 2.
From the equation of the ellipse,

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 6 Definite Integration Miscellaneous Exercise 6 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Miscellaneous Exercise 6

(I) Choose the correct alternative:

Question 1.
\(\int_{-9}^{9} \frac{x^{3}}{4-x^{2}} d x\) = ________
(a) 0
(b) 3
(c) 9
(d) -9
Answer:
(a) 0

Question 2.
\(\int_{-2}^{3} \frac{d x}{x+5}\) = _________
(a) -log(\(\frac{8}{3}\))
(b) log(\(\frac{8}{3}\))
(c) log(\(\frac{3}{8}\))
(d) -log(\(\frac{3}{8}\))
Answer:
(b) log(\(\frac{8}{3}\))

Question 3.
\(\int_{2}^{3} \frac{x}{x^{2}-1} d x\) = _________
(a) log(\(\frac{8}{3}\))
(b) -log(\(\frac{8}{3}\))
(c) \(\frac{1}{2}\) log(\(\frac{8}{3}\))
(d) \(\frac{-1}{2}\) log(\(\frac{8}{3}\))
Answer:
(c) \(\frac{1}{2}\) log(\(\frac{8}{3}\))

Question 4.
\(\int_{4}^{9} \frac{d x}{\sqrt{x}}\) = ___________
(a) 9
(b) 4
(c) 2
(d) 0
Answer:
(c) 2

Question 5.
If \(\int_{0}^{a} 3 x^{2} d x=8\), then a = __________
(a) 2
(b) 0
(c) \(\frac{8}{3}\)
(d) a
Answer:
(a) 2

Question 6.
\(\int_{2}^{3} x^{4}\) dx = ________
(a) \(\frac{1}{2}\)
(b) \(\frac{5}{2}\)
(c) \(\frac{5}{211}\)
(d) \(\frac{211}{5}\)
Answer:
(d) \(\frac{211}{5}\)

Question 7.
\(\int_{0}^{2} e^{x}\) dx = _______
(a) e – 1
(b) 1 – e
(c) 1 – e²
(d) e² – 1
Answer:
(d) e² – 1

Question 8.
\(\int_{a}^{b} f(x) d x\) = ________
(a) \(\int_{b}^{a} f(x) d x\)
(b) –\(\int_{a}^{b} f(x) d x\)
(c) –\(\int_{b}^{a} f(x) d x\)
(d) \(\int_{0}^{a} f(x) d x\)
Answer:
(c) –\(\int_{b}^{a} f(x) d x\)

Question 9.
\(\int_{-7}^{7} \frac{x^{3}}{x^{2}+7} d x\) = _________
(a) 7
(b) 49
(c) 0
(d) \(\frac{7}{2}\)
Answer:
(c) 0

Question 10.
\(\int_{2}^{7} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{9-x}} d x\) = _________
(a) \(\frac{7}{2}\)
(b) \(\frac{5}{2}\)
(c) 7
(d) 2
Answer:
(b) \(\frac{5}{2}\)

(II) Fill in the blanks:

Question 1.
\(\int_{0}^{2} e^{x} d x\) = ________
Answer:
e² – 1

Question 2.
\(\int_{2}^{3} x^{4} d x\) = __________
Answer:
\(\frac{211}{5}\)

Question 3.
\(\int_{0}^{1} \frac{d x}{2 x+5}\) = ____________
Answer:
\(\frac{1}{2} \log \left(\frac{7}{5}\right)\)

Question 4.
If \(\int_{0}^{a} 3 x^{2} d x\) = 8, then a = _________
Answer:
2

Question 5.
\(\int_{4}^{9} \frac{1}{\sqrt{x}} d x\) = _________
Answer:
2

Question 6.
\(\int_{2}^{3} \frac{x}{x^{2}-1} d x\) = _________
Answer:
\(\frac{1}{2} \log \left(\frac{8}{3}\right)\)

Question 7.
\(\int_{-2}^{3} \frac{d x}{x+5}\) = _________
Answer:
\(\log \left(\frac{8}{3}\right)\)

Question 8.
\(\int_{-9}^{9} \frac{x^{3}}{4-x^{2}} d x\) = _____________
Answer:
o

(III) State whether each of the following is True or False:

Question 1.
\(\int_{a}^{b} f(x) d x=\int_{-b}^{-a} f(x) d x\)
Answer:
True

Question 2.
\(\int_{a}^{b} f(x) d x=\int_{a}^{b} f(t) d t\)
Answer:
True

Question 3.
\(\int_{0}^{a} f(x) d x=\int_{a}^{0} f(a-x) d x\)
Answer:
False

Question 4.
\(\int_{a}^{b} f(x) d x=\int_{a}^{b} f(x-a-b) d x\)
Answer:
False

Question 5.
\(\int_{-5}^{5} \frac{x^{3}}{x^{2}+7} d x=0\)
Answer:
True

Question 6.
\(\int_{1}^{2} \frac{\sqrt{x}}{\sqrt{3-x}+\sqrt{x}} d x=\frac{1}{2}\)
Answer:
True

Question 7.
\(\int_{2}^{7} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{9-x}} d x=\frac{9}{2}\)
Answer:
False

Question 8.
\(\int_{4}^{7} \frac{(11-x)^{2}}{(11-x)^{2}+x^{2}} d x=\frac{3}{2}\)
Answer:
True

(IV) Solve the following:

Question 1.
\(\int_{2}^{3} \frac{x}{(x+2)(x+3)} d x\)
Solution:

Question 2.
\(\int_{1}^{2} \frac{x+3}{x(x+2)} d x\)
Solution:
Let I = \(\int_{1}^{2} \frac{x+3}{x(x+2)} d x\)
Let \(\frac{x+3}{x(x+2)}=\frac{A}{x}+\frac{B}{x+2}\)
∴ x + 3 = A(x + 2) + Bx
Put x = 0, we get
3 = A(2) + B(0)
∴ A = \(\frac{3}{2}\)
Put x + 2 = 0, i.e. x = -2, we get
-2 + 3 = A(0) + B(-2)
∴ 1 = -2B
∴ B = \(-\frac{1}{2}\)

Question 3.
\(\int_{1}^{3} x^{2} \log x d x\)
Solution:

Question 4.
\(\int_{0}^{1} e^{x^{2}} \cdot x^{3} d x\)
Solution:

Question 5.
\(\int_{1}^{2} e^{2 x}\left(\frac{1}{x}-\frac{1}{2 x^{2}}\right) d x\)
Solution:

Question 6.
\(\int_{4}^{9} \frac{1}{\sqrt{x}} d x\)
Solution:

Question 7.
\(\int_{-2}^{3} \frac{1}{x+5} d x\)
Solution:

Question 8.
\(\int_{2}^{3} \frac{x}{x^{2}-1} d x\)
Solution:

Question 9.
\(\int_{0}^{1} \frac{x^{2}+3 x+2}{\sqrt{x}} d x\)
Solution:

Question 10.
\(\int_{3}^{5} \frac{d x}{\sqrt{x+4}+\sqrt{x-2}}\)
Solution:

Question 11.
\(\int_{2}^{3} \frac{x}{x^{2}+1} d x\)
Solution:

Question 12.
\(\int_{1}^{2} x^{2} d x\)
Solution:

Question 13.
\(\int_{-4}^{-1} \frac{1}{x} d x\)
Solution:

Question 14.
\(\int_{0}^{1} \frac{1}{\sqrt{1+x}+\sqrt{x}} d x\)
Solution:

Question 15.
\(\int_{0}^{4} \frac{1}{\sqrt{x^{2}+2 x+3}} d x\)
Solution:

Question 16.
\(\int_{2}^{4} \frac{x}{x^{2}+1} d x\)
Solution:

Question 17.
\(\int_{0}^{1} \frac{1}{2 x-3} d x\)
Solution:

Question 18.
\(\int_{1}^{2} \frac{5 x^{2}}{x^{2}+4 x+3} d x\)
Solution:

Question 19.
\(\int_{1}^{2} \frac{d x}{x(1+\log x)^{2}}\)
Solution:

Question 20.
\(\int_{0}^{9} \frac{1}{1+\sqrt{x}} d x\)
Solution:

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 6 Definite Integration Ex 6.2 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Ex 6.2

Evaluate the following integrals:

Question 1.
\(\int_{-9}^{9} \frac{x^{3}}{4-x^{2}} d x\)
Solution:

Question 2.
\(\int_{0}^{a} x^{2}(a-x)^{3 / 2} d x\)
Solution:
We use the property

Question 3.
\(\int_{1}^{3} \frac{\sqrt[3]{x+5}}{\sqrt[3]{x+5}+\sqrt[3]{9-x}} d x\)
Solution:

Question 4.
\(\int_{2}^{5} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{7-x}} d x\)
Solution:

Question 5.
\(\int_{1}^{2} \frac{\sqrt{x}}{\sqrt{3-x}+\sqrt{x}} d x\)
Solution:

Question 6.
\(\int_{2}^{7} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{9-x}} d x\)
Solution:

Question 7.
\(\int_{0}^{1} \log \left(\frac{1}{x}-1\right) d x\)
Solution:

Question 8.
\(\int_{0}^{1} x(1-x)^{5} d x\)
Solution:
We use the property

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 6 Definite Integration Ex 6.1 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Ex 6.1

Evaluate the following definite integrals:

Question 1.
\(\int_{4}^{9} \frac{1}{\sqrt{x}} d x\)
Solution:

Question 2.
\(\int_{-2}^{3} \frac{1}{x+5} d x\)
Solution:

Question 3.
\(\int_{2}^{3} \frac{x}{x^{2}-1} d x\)
Solution:

Question 4.
\(\int_{0}^{1} \frac{x^{2}+3 x+2}{\sqrt{x}} d x\)
Solution:

Question 5.
\(\int_{2}^{3} \frac{x}{(x+2)(x+3)} d x\)
Solution:
Let I = \(\int_{2}^{3} \frac{x}{(x+2)(x+3)} d x\)
Let \(\frac{x}{(x+2)(x+3)}=\frac{A}{x+3}+\frac{B}{x+2}\)
∴ x = A(x + 2) + B(x + 3)
Put x + 3 = 0, i.e. x = -3, we get
-3 = A(-1) + B(0)
∴ A = 3
Put x + 2 = 0, i.e. x = -2, we get
-2 = A(0) + B(1)
∴ B = -2

Question 6.
\(\int_{1}^{2} \frac{d x}{x^{2}+6 x+5}\)
Solution:

Question 7.
If \(\int_{0}^{a}(2 x+1) d x\) = 2, find the real values of ‘a’.
Solution:
Let I = \(\int_{0}^{a}(2 x+1) d x\)
= \(\left[2 \cdot \frac{x^{2}}{2}+x\right]_{0}^{a}\)
= a² + a – 0
= a² + a
∴ I = 2 gives a² + a = 2
∴ a² + a – 2 = 0
∴ (a + 2)(a – 1) = 0 1
∴ a + 2 = 0 or a – 1 = 0
∴ a = -2 or a = 1.

Question 8.
If \(\int_{1}^{a}\left(3 x^{2}+2 x+1\right) d x\) = 11, find ‘a’.
Solution:
Let I = \(\int_{1}^{a}\left(3 x^{2}+2 x+1\right) d x\)
= \(\left[3\left(\frac{x^{3}}{3}\right)+2\left(\frac{x^{2}}{2}\right)+x\right]_{1}^{a}\)
= \(\left[x^{3}+x^{2}+x\right]_{1}^{a}\)
= (a³ + a² + a) – (1 + 1 + 1)
= a³ + a² + a – 3
∴ I = 11 gives a³ + a² + a – 3 = 11
∴ a³ + a² + a – 14 = 0
∴ (a³ – 8) + (a² + a – 6) = 0
∴ (a – 2)(a² + 2a + 4) + (a + 3)(a – 2) = 0
∴ (a – 2)(a² + 2a + 4 + a + 3) = 0
∴ (a – 2)(a² + 3a + 7) = 0
∴ a – 2 = 0 or a² + 3a + 7 = 0
∴ a = 2 or a = \(\frac{-3 \pm \sqrt{9-28}}{2}\)
The latter two roots are not real.
∴ they are rejected.
∴ a = 2.

Question 9.
\(\int_{0}^{1} \frac{1}{\sqrt{1+x}+\sqrt{x}} d x\)
Solution:

Question 10.
\(\int_{1}^{2} \frac{3 x}{9 x^{2}-1} d x\)
Solution:
Let I = \(\int_{1}^{2} \frac{3 x}{9 x^{2}-1} d x\) = \(\int_{1}^{2} \frac{3 x}{(3 x)^{2}-1} d x\)
Put 3x = t
∴ 3 dx = dt
∴ dx = \(\frac{d t}{3}\)
When x = 1, t = 3 × 1 = 3
When x = 2, t = 3 × 2 = 6

Question 11.
\(\int_{1}^{3} \log x d x\)
Solution:

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 5 Integration Miscellaneous Exercise 5 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 5 Integration Miscellaneous Exercise 5

(I) Choose the correct alternative from the following:

Question 1.
The value of \(\int \frac{d x}{\sqrt{1-x}}\) is
(a) 20\(\sqrt{1-x}\) + c
(b) -2\(\sqrt{1-x}\) + c
(c) √x + c
(d) x + c
Answer:
(b) -2\(\sqrt{1-x}\) + c

Question 2.
\(\int \sqrt{1+x^{2}} d x\) =
(a) \(\frac{x}{2} \sqrt{1+x^{2}}+\frac{1}{2} \log \left(x+\sqrt{1+x^{2}}\right)+c\)
(b) \(\frac{2}{3}\left(1+x^{2}\right)^{3 / 2}+c\)
(c) \(\frac{1}{3}\left(1+x^{2}\right)+c\)
(d) \(\frac{(x)}{\sqrt{1+x^{2}}}+c\)
Answer:
(a) \(\frac{x}{2} \sqrt{1+x^{2}}+\frac{1}{2} \log \left(x+\sqrt{1+x^{2}}\right)+c\)

Question 3.
\(\int x^{2}(3)^{x^{3}} d x\) =
(a) \(\text { (3) }^{x^{3}}+c\)
(b) \(\frac{(3)^{x^{3}}}{3 \cdot \log 3}+c\)
(c) \(\log 3(3)^{x^{3}}+c\)
(d) \(x^{2}(3)^{x 3}\)
Answer:
(b) \(\frac{(3)^{x^{3}}}{3 \cdot \log 3}+c\)
Hint:
Put x³ = t

Question 4.
\(\int \frac{x+2}{2 x^{2}+6 x+5} d x=p \int \frac{4 x+6}{2 x^{2}+6 x+5} d x\) + \(\frac{1}{2} \int \frac{d x}{2 x^{2}+6 x+5}\), then p = __________
(a) \(\frac{1}{3}\)
(b) \(\frac{1}{2}\)
(c) \(\frac{1}{4}\)
(d) 2
Answer:
(c) \(\frac{1}{4}\)
Hint:
\(\int \frac{x+2}{2 x^{2}+6 x+5} d x=\int \frac{\frac{1}{4}(4 x+6)+\frac{1}{2}}{2 x^{2}+6 x+5} d x\)

Question 5.
\(\int \frac{d x}{\left(x-x^{2}\right)}\) = ________
(a) log x – log(1 – x) + c
(b) log(1 – x²) + c
(c) -log x + log(1 – x) + c
(d) log(x – x²) + c
Answer:
(a) log x – log(1 – x) + c

Question 6.
\(\int \frac{d x}{(x-8)(x+7)}\) = __________
(a) \(\frac{1}{15} \log \left|\frac{x+2}{x-1}\right|+c\)
(b) \(\frac{1}{15} \log \left|\frac{x+8}{x+7}\right|+c\)
(c) \(\frac{1}{15} \log \left|\frac{x-8}{x+7}\right|+c\)
(d) (x – 8)(x – 7) + c
Answer:
(c) \(\frac{1}{15} \log \left|\frac{x-8}{x+7}\right|+c\)

Question 7.
\(\int\left(x+\frac{1}{x}\right)^{3} d x\) = _________
(a) \(\frac{1}{4}\left(x+\frac{1}{x}\right)^{4}+c\)
(b) \(\frac{x^{4}}{4}+\frac{3 x^{2}}{2}+3 \log x-\frac{1}{2 x^{2}}+c\)
(c) \(\frac{x^{4}}{4}+\frac{3 x^{2}}{2}+3 \log x+\frac{1}{x^{2}}+c\)
(d) \(\left(x-x^{-1}\right)^{3}+c\)
Answer:
(b) \(\frac{x^{4}}{4}+\frac{3 x^{2}}{2}+3 \log x-\frac{1}{2 x^{2}}+c\)
Hint:
\(\left(x+\frac{1}{x}\right)^{3}=x^{3}+3 x+\frac{3}{x}+\frac{1}{x^{3}}\)

Question 8.
\(\int\left(\frac{e^{2 x}+e^{-2 x}}{e^{x}}\right) d x\)
(a) \(e^{x}-\frac{1}{3 e^{3 x}}+c\)
(b) \(e^{x}+\frac{1}{3 e^{3 x}}+c\)
(c) \(e^{-x}+\frac{1}{3 e^{3 x}}+c\)
(d) \(e^{-x}-\frac{1}{3 e^{3 x}}+c\)
Answer:
(a) \(e^{x}-\frac{1}{3 e^{3 x}}+c\)

Question 9.
∫(1 – x)^-2 dx = ___________
(a) (1 + x)^-1 + c
(b) (1 – x)^-1 + c
(c) (1 – x)^-1 – 1 + c
(d) (1 – x)^-1 + 1 + c
Answer:
(b) (1 – x)^-1 + c

Question 10.
\(\int \frac{\left(x^{3}+3 x^{2}+3 x+1\right)}{(x+1)^{5}} d x\) = _______
(a) \(\frac{-1}{x+1}+c\)
(b) \(\left(\frac{-1}{x+1}\right)^{5}+c\)
(c) log(x + 1) + c
(d) log|x + 1|⁵ + c
Answer:
(a) \(\frac{-1}{x+1}+c\)
Hint:
x³ + 3x² + 3x + 1 = (x + 1)³

(II) Fill in the blanks.

Question 1.
\(\int \frac{5\left(x^{6}+1\right)}{x^{2}+1}\)dx = x⁴ + ___x³ + 5x + c.
Answer:
\(-\frac{5}{3}\)
Hint:
x⁶ + 1 = (x² + 1)(x⁴ – x² + 1)

Question 2.
\(\int \frac{x^{2}+x-6}{(x-2)(x-1)} d x\) = x + ______ + c
Answer:
4 log|x – 1|
Hint:
x² + x – 6 = (x + 3)(x – 2)

Question 3.
If f'(x) = \(\frac{1}{x}\) + x and f(1) = \(\frac{5}{2}\) then f(x) = log x + \(\frac{x^{2}}{2}\) + _______
Answer:
2
Hint:

Question 4.
To find the value of \(\int \frac{(1+\log x) d x}{x}\) the proper substitution is __________
Answer:
1 + log x = t

Question 5.
\(\int \frac{1}{x^{3}}\left[\log x^{x}\right]^{2} d x\) = p(log x)³ + c, then p = _______
Answer:
\(\frac{1}{3}\)
Hint:
\(\frac{1}{x^{3}}\left(\log x^{x}\right)^{2}=\frac{1}{x^{3}}(x \log x)^{2}=\frac{(\log x)^{2}}{x}\)

(III) State whether each of the following is True or False:

Question 1.
The proper substitution for \(\int x\left(x^{x}\right)^{x}(2 \log x+1) d x \text { is }\left(x^{x}\right)^{x}=t\)
Answer:
True

Question 2.
If ∫x e^2x dx is equal to e^2x f(x) + c where c is constant of integration, then f(x) is \(\frac{(2 x-1)}{2}\).
Answer:
False

Question 3.
If ∫x f(x) dx = \(\frac{f(x)}{2}\), then f(x) = \(e^{x^{2}}\).
Answer:
True

Question 4.
If \(\int \frac{(x-1) d x}{(x+1)(x-2)}\) = A log|x + 1| + B log|x – 2|, then A + B = 1.
Answer:
True

Question 5.
For \(\int \frac{x-1}{(x+1)^{3}} e^{x} d x\) = e^x f(x) + c, f(x) = (x + 1)².
Answer:
False

(IV) Solve the following:

1. Evaluate:

(i) \(\int \frac{5 x^{2}-6 x+3}{2 x-3} d x\)
Solution:

(ii) \(\int(5 x+1)^{\frac{4}{9}} d x\)
Solution:

(iii) \(\int \frac{1}{2 x+3} d x\)
Solution:

(iv) \(\int \frac{x-1}{\sqrt{x+4}} d x\)
Solution:

(v) If f'(x) = √x and f(1) = 2, then find the value of f(x).
Solution:
By the definition of integral

(vi) ∫|x| dx if x < 0
Solution:
∫|x| dx = ∫-x dx …..[∵ x < 0]
= -∫x dx
= \(-\frac{x^{2}}{2}\) + c

2. Evaluate:

(i) Find the primitive of \(\frac{1}{1+e^{x}}\)
Solution:
Let I be the primitive of \(\frac{1}{1+e^{x}}\)

(ii) \(\int \frac{a e^{x}+b e^{-x}}{\left(a e^{x}-b e^{-x}\right)^{2}} d x\)
Solution:

(iii) \(\int \frac{1}{2 x+3 x \log x} d x\)
Solution:

(iv) \(\int \frac{1}{\sqrt{x}+x} d x\)
Solution:

(v) \(\int \frac{2 e^{x}-3}{4 e^{x}+1} d x\)
Solution:

3. Evaluate:

(i) \(\int \frac{d x}{\sqrt{4 x^{2}-5}} d x\)
Solution:

(ii) \(\int \frac{d x}{3-2 x-x^{2}} d x\)
Solution:

(iii) \(\int \frac{d x}{9 x^{2}-25}\)
Solution:

(iv) \(\int \frac{e^{x}}{\sqrt{e^{2 x}+4 e^{x}+13}} d x\)
Solution:

(v) \(\int \frac{d x}{x\left[(\log x)^{2}+4 \log x-1\right]}\)
Solution:

(vi) \(\int \frac{d x}{5-16 x^{2}}\)
Solution:

(vii) \(\int \frac{d x}{25 x-x(\log x)^{2}}\)
Solution:

(viii) \(\int \frac{e^{x}}{4 e^{2 x}-1} d x\)
Solution:

4. Evaluate:

(i) ∫(log x)² dx
Solution:

(ii) \(\int e^{x} \frac{1+x}{(2+x)^{2}} d x\)
Solution:

(iii) ∫x e^2x dx
Solution:

(iv) ∫log(x² + x) dx
Solution:

(v) \(\int e^{\sqrt{x}} d x\)
Solution:

(vi) \(\int \sqrt{x^{2}+2 x+5} d x\)
Solution:

(vii) \(\int \sqrt{x^{2}-8 x+7} d x\)
Solution:

5. Evaluate:

(i) \(\int \frac{3 x-1}{2 x^{2}-x-1} d x\)
Solution:

(ii) \(\int \frac{2 x^{3}-3 x^{2}-9 x+1}{2 x^{2}-x-10} d x\)
Solution:

(iii) \(\int \frac{1+\log x}{x(3+\log x)(2+3 \log x)} d x\)
Solution:

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 5 Integration Ex 5.6 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 5 Integration Ex 5.6

Evaluate:

Question 1.
\(\int \frac{2 x+1}{(x+1)(x-2)} d x\)
Solution:
Let I = \(\int \frac{2 x+1}{(x+1)(x-2)} d x\)
Let \(\frac{2 x+1}{(x+1)(x-2)}=\frac{A}{x+1}+\frac{B}{x-2}\)
∴ 2x + 1 = A(x – 2) + B(x + 1)
Put x + 1 = 0, i.e. x = -1, we get
2(-1) + 1 = A(-3) + B(0)
∴ A = \(\frac{1}{3}\)
Put x – 2 = 0, i.e. x = 2, we get
2(2) + 1 = A(0) + B(3)
∴ B = \(\frac{5}{3}\)

Question 2.
\(\int \frac{2 x+1}{x(x-1)(x-4)} d x\)
Solution:
Let I = \(\int \frac{2 x+1}{x(x-1)(x-4)} d x\)
Let \(\int \frac{2 x+1}{x(x-1)(x-4)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x-4}\)
∴ 2x + 1 = A(x – 1)(x – 4) + Bx(x – 4) + Cx(x – 1)
Put x = 0, we get
2(0) + 1 = A(-1)(-4) + B(0)(-4) + C(0)(-1)
∴ 1 = 4A
∴ A = \(\frac{1}{4}\)
Put x – 1 = 0, i.e. x = 1, we get
2(1) + 1 = A(0)(-3) + B(1)(-3) + C(1)(0)
∴ 3 = -3B
∴ B = -1
Put x – 4 = 0, i.e x = 4, we get
2(4) + 1 = A(3)(0) + B(4)(0) + C(4)(3)
∴ 9 = 12C
∴ C = \(\frac{3}{4}\)

Question 3.
\(\int \frac{x^{2}+x-1}{x^{2}+x-6} d x\)
Solution:

∴ 1 = A(x – 2) + B(x + 3)
Put x + 3 = 0, i.e. x = -3, we get
1 = A(-5) + B (0)
∴ A = \(\frac{-1}{5}\)
Put x – 2 = 0, i.e. x = 2, we get
1 = A(0) + B(5)
∴ B = \(\frac{1}{5}\)

Question 4.
\(\int \frac{x}{(x-1)^{2}(x+2)} d x\)
Solution:
Let I = \(\int \frac{x}{(x-1)^{2}(x+2)} d x\)
Let \(\frac{x}{(x-1)^{2}(x+2)}=\frac{A}{x-1}+\frac{B}{(x-1)^{2}}+\frac{C}{x+2}\)
∴ x = A(x – 1)(x + 2) + B(x + 2) + C(x – 1)²
Put x – 1 = 0, i.e. x = 1, we get
1 = A(0)(3) + B(3) + C(0)
∴ B = \(\frac{1}{3}\)
Put x + 2 = 0, i.e. x = -2, we get
-2 = A (-3)(0) + B(0) + C(9)
∴ C = \(-\frac{2}{9}\)
Put x = -1, we get,
-1 = A(-2)(1) + B(1) + C(4)
But B = \(\frac{1}{3}\) and C = \(-\frac{2}{9}\)

Question 5.
\(\int \frac{3 x-2}{(x+1)^{2}(x+3)} d x\)
Solution:
Let I = \(\int \frac{3 x-2}{(x+1)^{2}(x+3)} d x\)
Let \(\frac{3 x-2}{(x+1)^{2}(x+3)}=\frac{A}{x+1}+\frac{B}{(x+1)^{2}}+\frac{C}{x+3}\)
∴ 3x – 2 = A(x + 1)(x + 3) + B(x + 3) + C(x + 1)²
Put x + 1 = 0, i.e. x = -1, we get
3(-1) – 2 = A(0)(2) + B(2) + C(0)
∴ -5 = 2B
∴ B = \(-\frac{5}{2}\)
Put x + 3 = 0, i.e. x = -3, we get
3(-3) – 2 = A(-2)(0) + B(0) + C(4)
∴ -11 = 4C
∴ C = \(-\frac{11}{4}\)
Put x = 0, we get
3(0) – 2 = A(1)(3) + B(3) + C(1)
∴ -2 = 3A + 3B + C
But B = \(-\frac{5}{2}\) and C = \(-\frac{11}{4}\)

Question 6.
\(\int \frac{1}{x\left(x^{5}+1\right)} d x\)
Solution:
Let I = \(\int \frac{1}{x\left(x^{5}+1\right)} d x\)
= \(\int \frac{x^{4}}{x^{5}\left(x^{5}+1\right)} d x\)

Question 7.
\(\int \frac{1}{x\left(x^{n}+1\right)} d x\)
Solution:

Question 8.
\(\int \frac{5 x^{2}+20 x+6}{x^{3}+2 x^{2}+x} d x\)
Solution:

Let \(\frac{5 x^{2}+20 x+6}{x(x+1)^{2}}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{(x+1)^{2}}\)
∴ 5x² + 20x + 6 = A(x + 1)² + Bx(x + 1) + Cx
Put x = 0, we get
0 + 0 + 6 = A(1) + B(0)(1) + C(0)
∴ A = 6
Put x + 1 = 0, i.e. x = -1, we get
5(1) + 20(-1) + 6 = A(0) + B(-1)(0) + C(-1)
∴ -9 = -C
∴ C = 9
Put x = 1, we get
5(1) + 20(1) + 6 = A(4) + B(1)(2) + C(1)
But A = 6 and C = 9
∴ 31 = 24 + 2B + 9
∴ B = -1

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 5 Integration Ex 5.5 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 5 Integration Ex 5.5

Evaluate the following.

Question 1.
∫x log x
Solution:

Question 2.
∫x² e^4x dx
Solution:

Question 3.
∫x² e^3x dx
Solution:

Question 4.
\(\int x^{3} e^{x^{2}} d x\)
Solution:

Question 5.
\(\int e^{x}\left(\frac{1}{x}-\frac{1}{x^{2}}\right) d x\)
Solution:

Question 6.
\(\int e^{x} \frac{x}{(x+1)^{2}} d x\)
Solution:

Question 7.
\(\int e^{x} \frac{x-1}{(x+1)^{3}} d x\)
Solution:

Question 8.
\(\int e^{x}\left[(\log x)^{2}+\frac{2 \log x}{x}\right] d x\)
Solution:

Question 9.
\(\int\left[\frac{1}{\log x}-\frac{1}{(\log x)^{2}}\right] d x\)
Solution:

Question 10.
\(\int \frac{\log x}{(1+\log x)^{2}} d x\)
Solution:

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 5 Integration Ex 5.4 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 5 Integration Ex 5.4

Evaluate the following.

Question 1.
\(\int \frac{1}{4 x^{2}-1} d x\)
Solution:

Question 2.
\(\int \frac{1}{x^{2}+4 x-5} d x\)
Solution:

Question 3.
\(\int \frac{1}{4 x^{2}-20 x+17} d x\)
Solution:

Question 4.
\(\int \frac{x}{4 x^{4}-20 x^{2}-3} d x\)
Solution:

Question 5.
\(\int \frac{x^{3}}{16 x^{8}-25} d x\)
Solution:

Question 6.
\(\int \frac{1}{a^{2}-b^{2} x^{2}} d x\)
Solution:

Question 7.
\(\int \frac{1}{7+6 x-x^{2}} d x\)
Solution:

Question 8.
\(\int \frac{1}{\sqrt{3 x^{2}+8}} d x\)
Solution:

Question 9.
\(\int \frac{1}{\sqrt{x^{2}+4 x+29}} d x\)
Solution:

Question 10.
\(\int \frac{1}{\sqrt{3 x^{2}-5}} d x\)
Solution:

Question 11.
\(\int \frac{1}{\sqrt{x^{2}-8 x-20}} d x\)
Solution:

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 5 Integration Ex 5.3 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 5 Integration Ex 5.3

Evaluate the following:

Question 1.
\(\int \frac{3 e^{2 t}+5}{4 e^{2 t}-5} d t\)
Solution:
Let I = \(\int \frac{3 e^{2 t}+5}{4 e^{2 t}-5} d t\)
Put, Numerator = A(Denominator) + B[\(\frac{d}{d x}\)(Denominator)]
∴ 3e^2t + 5 = A(4e^2t – 5) + B[\(\frac{d}{d t}\)(4e^2t – 5)]
∴ 3e^2t + 5 = A(4e^2t – 5) + B[4e^2t × 2 – 0]
∴ 3e^2t + 5 = (4A + 8B) e^2t – 5A
Equating the coefficient of e^2t and constant on both sides, we get
4A + 8B = 3
and -5A = 5
∴ A = -1
∴ from (1), 4(-1) + 8B = 3
∴ 8B = 7
∴ B = \(\frac{7}{8}\)

Question 2.
\(\int \frac{20-12 e^{x}}{3 e^{x}-4} d x\)
Solution:
Let I = \(\int \frac{20-12 e^{x}}{3 e^{x}-4} d x\)
Put, Numerator = A (Denominator) + B[\(\frac{d}{d x}\)(Denominator)]
∴ 20 – 12e^x = A(3e^x – 4) + B[\(\frac{d}{d x}\)(3e^x – 4)]
∴ 20 – 12e^x = A(3e^x – 4) + B(3e^x – 0)
∴ 20 – 12e^x = (3A + 3B)e^x – 4A
Equating the coefficient of ex and constant on both sides, we get
3A + 3B = -12 ……(1)
and -4A = 20
∴ A = -5
from (1), 3(-5) + 3B = -12
∴ 3B = 3
∴ B = 1
∴ 20 – 12e^x = -5(3e^x – 4) + (3e^x)

Question 3.
\(\int \frac{3 e^{x}+4}{2 e^{x}-8} d x\)
Solution:
Let I = \(\int \frac{3 e^{x}+4}{2 e^{x}-8} d x\)
Put, Numerator = A (Denominator) + B[\(\frac{d}{d x}\)(Denominator)]
∴ 3e^x + 4 = A(2e^x – 8) + B[\(\frac{d}{d x}\)(2e^x – 8)]
∴ 3e^x + 4 = A(2e^x – 8) + B(2e^x – 0)
∴ 3e^x + 4 = (2A + 2B)e^x – 8A
Equating the coefficient of ex and constant on both sides, we get
2A + 2B = 3 ……..(1)
and -8A = 4
∴ A = \(-\frac{1}{2}\)
∴ from (1), 2(\(-\frac{1}{2}\)) + 2B = 3
∴ 2B = 4
∴ B = 2

Question 4.
\(\int \frac{2 e^{x}+5}{2 e^{x}+1} d x\)
Solution: