Samacheer Class 12 Maths Volume 2 Applications of Integration Book Back Answers & Solutions

Q: Evaluate the following definite integrals. (i) \int_{3}^{4} \frac{dx}{x^2-4}

(iv) \int_{0}^{\pi / 2} (\frac{1+sin}{1+cosx})dx (v) \int_{0}^{\pi / 2} \sqrt{cos θ} sin³θ dθ (vi) \int_{0}^{1} \frac{1-x^2}{(1+x^2)^2} dx

Q: Evaluate the following: (i) \int_{0}^{π/2} \frac{dx}{1+5cos^2x}

(ii) \int_{0}^{π/2} \frac{dx}{5+4sin^2x}

Q: Evaluate the following (i) \int_{0}^{π/2} sin 10 x dx

Aliter Method (viii) \int_{1}^{0} x² (1 – x)³ dx By applying reduction formula

Q: Evaluate the following (i) \int_{0}^{∞} x 5 e -3x dx

(ii) \int_{0}^{π/2} \frac{e^{-tanx}}{cos^6 x} dx

Q: Find the area of the region bounded by 3x – 2y + 6 = 0, x = -3, x = 1 and x axis.

3x – 2y + 6 = 0 2y = 3x + 6 y = \frac { 1 }{ 2 }(3x + 6)

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evaluationevaluation2 questions

Q.1Explain the right-end and mid-point Riemann rules and remarks about geometric interpretation of the Riemann integral.v

Solution

Concise statement of the rules and geometric remarks as given above.

Answer:

Right-end rule: choose sample points ξ_i = x_i (right endpoints); then ∫_a^b f(x) dx = lim_{n→∞} Σ_{i=1}^n f(x_i)Δx_i when limit exists. Mid-point rule: choose ξ_i = (x_{i-1}+x_i)/2; then ∫_a^b f(x) dx = lim_{n→∞} Σ f(midpoint)Δx_i. Remarks: (1) F(x)=∫_a^x f(u) du defines an antiderivative. (2) If f≥0 on [a,b], the integral equals the geometric area under the curve. (3) If f≤0 the integral equals negative of the geometric area. (4) If f changes sign, split [a,b] into subintervals where f has constant sign and add/subtract areas accordingly.

Q.2State Bernoulli's formula for integration by parts repeated application.v

Solution

Repeated integration by parts produces the series u v − u' v_1 + u'' v_2 − … until derivatives vanish; this is called Bernoulli's formula for products.

Answer:

Bernoulli's formula: ∫ u dv = uv − ∫ v du; applying repeatedly gives ∫ u dv = uv − u'v_1 + u''v_2 − ... alternating signs, where v_k are successive antiderivatives of v.

EXERCISE 9.1EXERCISE 9.13 questions

Q.3Find an approximate value of ∫_1^5 x dx by applying the left-end rule with partition {1,2,3,4,5}.v

Solution

Δx =1. Left endpoints: 1,2,3,4. Left Riemann sum = (1+2+3+4)·1 = 10.

Answer:

Q.4Find an approximate value of ∫_1^5 x dx by applying the right-end rule with partition {1,2,3,4,5}.v

Solution

Δx =1. Right endpoints: 2,3,4,5. Right Riemann sum = (2+3+4+5)·1 = 14.

Answer:

Q.5Find an approximate value of ∫_1^5 x dx by applying the mid-point rule with partition {1,2,3,4,5}.v

Solution

Δx =1. Midpoints: 1.5,2.5,3.5,4.5. Sum = (1.5+2.5+3.5+4.5)·1 = 12, which equals the exact value ∫_1^5 x dx = (25−1)/2 =12.

Answer:

EXERCISE 9.2EXERCISE 9.21 questions

Q.6Evaluate the following integrals as limits of sums: (i) ∫_4^5 (x+5) dx (ii) ∫_1^4 (2−x) dxv

Solution

(i) ∫_4^5 (x+5) dx = [x^2/2 +5x]_4^5 = (25/2+25)−(16/2+20) = 19/2. (ii) ∫_1^4 (2−x) dx = [2x − x^2/2]_1^4 = (8−8)−(2−1/2) = −3/2.

Answer:

(i) 19/2 (ii) −3/2

EXERCISE 9.3EXERCISE 9.32 questions

Q.7Evaluate the following definite integrals.
(i) $\int_{3}^{4}$ $\frac{dx}{x^2-4}$v

Solution

(ii) $\int_{-1}^{1}$ $\frac{dx}{x^2+2x+5}$

(iii) $\int_{0}^{1}$ $\frac{\sqrt{1-x}}{\sqrt{1+x}}$ dx

(iv) $\int_{0}^{\pi / 2}$ ($\frac{1+sin}{1+cosx}$)dx

(v) $\int_{0}^{\pi / 2}$ $\sqrt{cos θ}$ sin³θ dθ

(vi) $\int_{0}^{1}$ $\frac{1-x^2}{(1+x^2)^2}$ dx

Answer:

(iv) $\int_{0}^{\pi / 2}$ ($\frac{1+sin}{1+cosx}$)dx (v) $\int_{0}^{\pi / 2}$ $\sqrt{cos θ}$ sin³θ dθ (vi) $\int_{0}^{1}$ $\frac{1-x^2}{(1+x^2)^2}$ dx

Q.8Evaluate the following integrals using properties of integration:
(i) $\int_{-5}^{5}$ x cos ($\frac{e^x-1}{e^x+1}$) dxv

Solution

(ii) $\int_{-\pi / 2}^{\pi / 2}$ (x 5 + x cos x + tan³ x) dx
$\int_{-\pi / 2}^{\pi / 2}$ (x 5 + x cos x + tan³ x) dx
= $\int_{-\pi / 2}^{\pi / 2}$ (x 5 + x cos x + tan³ x) dx + $\int_{-\pi / 2}^{\pi / 2}$
Let f(x) = x 5 + x cos x + tan³x
f(-x) = -x 5 – x cos x – tan³x
f(x) = -f(-x)
f(x) is an odd function

(iii) $\int_{-\pi / 4}^{\pi / 4}$ sin² x dx
I = $\int_{-\pi / 4}^{\pi / 4}$ sin² x dx
f(x) = sin²x
f(-x) = sin²(-x) = sin²x
f(x) = f(-x)
f(x) is an even function

(iv) $\int_{0}^{2π}$ x log($\frac{3+cos x}{3-cos x}$)dx

(v) $\int_{0}^{π}$ sin 4 x cos³ x dx
$\int_{0}^{π}$ sin 4 x cos³ x dx
f(x) = sin 4 x cos³x
f(2π – x) = sin 4 (2π – x) cos³ (2π – x)
= sin 4 x cos³x
f(2π – x) = f(x)

Limit from 0 to π tends to 0 to 0
∴ Integral value = 0
∴ $\int_{0}^{π}$ sin 4 x cos³ x dx = 0
(vi) $\int_{0}^{1}$ |5x – 3|dx

(vii) $\int_{0}^{sin^2x}$ sin -1 √t dt + $\int_{0}^{cos^2x}$ cos -1 √t dt
I 1 = $\int_{0}^{sin^2x}$ sin -1 √t dt
Put sin -1 √t = θ
√t = sin θ
$\frac{1}{2√t}$ dt = cos θ dθ
dt = 2√t cos θ dθ
= 2 sin θ cos θ dθ
dt = sin 2θ dθ

I 1 = $\int_{0}^{cos^2x}$ cos -1 √t dt
Put cos -1 √t = θ
√t = cos θ
$\frac{1}{2√t}$ dt = -sin θ dθ
dt = -2√t sin θ dθ
= -2 cos θ sin θ dθ
dt = -sin 2θ dθ

(viii) $\int_{0}^{1}$ $\frac{log(1+x)}{1+x^2}$ dx

(ix) $\int_{0}^{π}$ $\frac{x sin x}{1+sin x}$ dx

(x) $\int_{π/8}^{3π/8}$ $\frac{1}{1+\sqrt{tan x}}$ dx

(xi) $\int_{0}^{π}$ x[sin²(sin x) + cos² (cos x)] dx
Let I = $\int_{0}^{π}$ x[sin²(sin x) + cos² (cos x)] dx
f(x) = sin² (sin x) + cos² (cos x)
f(π – x) = sin² (sin π – x)) + cos² (cos(π – x))
= sin² (sin x) + cos² (cos x)
f(x) = f(π – x)

Answer:

f(π – x) = sin² (sin π – x)) + cos² (cos(π – x)) = sin² (sin x) + cos² (cos x) f(x) = f(π – x)

EXERCISE 9.4EXERCISE 9.42 questions

Q.9Evaluate the following
$\int_{0}^{1}$ x³e -2x dxv

Solution

Bernoulli’s formula,
∫uv dx = uv 1 – u 1 v 2 + u 2 v 3 – u 3 v 4 + ….

Answer:

Bernoulli’s formula, ∫uv dx = uv 1 – u 1 v 2 + u 2 v 3 – u 3 v 4 + ….

Q.10Evaluate ∫_0^π x^2 cos(2x) dx.v

Solution

Let I=∫_0^π x^2 cos2x dx. Integrate by parts: u=x^2, dv=cos2x dx ⇒ v=½ sin2x. So I = [½ x^2 sin2x]_0^π − ∫_0^π x sin2x dx. The boundary term is 0. For J=∫_0^π x sin2x dx use parts: u=x, dv=sin2x dx ⇒ v=−½ cos2x, so J = [−½ x cos2x]_0^π + ½ ∫_0^π cos2x dx = −(π/2)·1 + 0 = −π/2. Hence I = −J = π/2.

Answer:

π/2

EXERCISE 9.5EXERCISE 9.51 questions

Q.11Evaluate the following:
(i) $\int_{0}^{π/2}$ $\frac{dx}{1+5cos^2x}$v

Solution

(ii) $\int_{0}^{π/2}$ $\frac{dx}{5+4sin^2x}$

Answer:

(ii) $\int_{0}^{π/2}$ $\frac{dx}{5+4sin^2x}$

EXERCISE 9.6EXERCISE 9.61 questions

Q.12Evaluate the following
(i) $\int_{0}^{π/2}$ sin 10 x dxv

Solution

Here n = 10, which is even

(ii) $\int_{0}^{π/2}$ cos 7 x dx
Here n = 7, which is odd

(iii) $\int_{0}^{π/4}$ sin 6 2x dx
put t = 2x
dt = 2 dx

(iv) $\int_{0}^{π/6}$ sin 5 3x dx
put t = 3x
dt = 3 dx

(v) $\int_{0}^{π/2}$ sin 2 x cos 4 x dx
Here m = 2, which is even and n = 4, which is even

(vi) $\int_{0}^{2π}$ sin 7 $\frac{x}{4}$ dx

(vii) $\int_{0}^{π/2}$ sin 3 θ cos 5 θ dθ

Aliter Method

(viii) $\int_{1}^{0}$ x² (1 – x)³ dx
By applying reduction formula

Answer:

Aliter Method (viii) $\int_{1}^{0}$ x² (1 – x)³ dx By applying reduction formula

EXERCISE 9.7EXERCISE 9.72 questions

Q.13Evaluate the following
(i) $\int_{0}^{∞}$ x 5 e -3x dxv

Solution

(ii) $\int_{0}^{π/2}$ $\frac{e^{-tanx}}{cos^6 x}$ dx

Answer:

(ii) $\int_{0}^{π/2}$ $\frac{e^{-tanx}}{cos^6 x}$ dx

Q.14$\int_{0}^{∞}$ sin αx² x³ dx = 32, α > 0, find αv

Solution

Answer:

See the worked solution above.

EXERCISE 9.8EXERCISE 9.811 questions

Q.15Find the area of the region bounded by 3x – 2y + 6 = 0, x = -3, x = 1 and x axis.v

Solution

3x – 2y + 6 = 0
2y = 3x + 6
y = $\frac { 1 }{ 2 }$(3x + 6)

Answer:

3x – 2y + 6 = 0 2y = 3x + 6 y = $\frac { 1 }{ 2 }$(3x + 6)

Q.16Find the area of the region bounded by 2x – y + 1 = 0, y = -1, y = 3 and y axis.v

Solution

Given straight line is 2x – y + 1 = 0
y = 2x + 1, x = $\frac { y-1 }{ 2 }$

= 1 + 1
= 2 sq. units
Area required = 2 sq. units

Answer:

= 1 + 1 = 2 sq. units Area required = 2 sq. units

Q.17The curve y = (x – 2)² + 1 has a minimum point at P. A point Q on the curve is such that the slope of PQ is 2. Find the area bounded by the curve and the chord PQ.v

Solution

Given curve is y = (x – 2)² + 1
(i.e) (y – 1) = (x – 2)²
Vertex of the parabola is (2, 1)
Minimum point P is (2, 1)
Slope of PQ is 2.

Equation of PQ is y – 1 = 2 (x – 2)
y – 1 = 2x – 4
y = 2x – 3
Intersecting point of y = 2x – 3 and y = (x – 2)² + 1
2x – 3 = (x – 2)² + 1
2x – 4 = (x – 2)²
2(x – 2) = (x – 2)²
x – 2 = 2
x = 4
when x = 4, y = 5

Required Area = $\frac { 4 }{ 3 }$ sq. units

Answer:

x = 4 when x = 4, y = 5 Required Area = $\frac { 4 }{ 3 }$ sq. units

Q.18Find the area of the region bounded by the curve 2 + x – x² + y = 0, x axis, x = -3 and x = 3v

Solution

Given curve is
2 + x – x² + y = 0
y = x² – x – 2
y = (x – 2)(x + 1)

Answer:

2 + x – x² + y = 0 y = x² – x – 2 y = (x – 2)(x + 1)

Q.19Find the area of the region bounded by the line $y=2x+5$ and the parabola $y=x^2-2x$.v

Solution

The curves meet where $x^2-2x=2x+5$, i.e. $x^2-4x-5=0\Rightarrow(x-5)(x+1)=0$, so $x=-1$ and $x=5$. On $[-1,5]$ the line lies above the parabola, so the area is $\displaystyle\int_{-1}^{5}\big[(2x+5)-(x^2-2x)\big]dx=\int_{-1}^{5}(-x^2+4x+5)\,dx=\left[-\dfrac{x^3}{3}+2x^2+5x\right]_{-1}^{5}=\dfrac{100}{3}-\left(-\dfrac{8}{3}\right)=36.$

Answer:

Area $=36$ square units.

Q.20Find the area between y = sin x and y = cos x for x ∈ [0,π].v

Solution

Solve sin x = cos x ⇒ x = π/4 in [0,π]. Area = ∫_0^{π/4} (cos x − sin x) dx + ∫_{π/4}^{π} (sin x − cos x) dx = [sin x + cos x]_0^{π/4} + [−cos x − sin x]_{π/4}^{π} = (√2−1)+(1+√2)=2√2.

Answer:

2√2

Q.21Find the area of the region bounded by y = tan x, y = cot x and the lines x = 0, x = π/2, y = 0.v

Solution

For 0 ≤ x ≤ π/4, the region above y=0 and below y=tan x contributes A1 = ∫_0^{π/4} tan x dx = [-ln|cos x|]_0^{π/4} = ln √2 = (1/2)ln2. For π/4 ≤ x ≤ π/2 the region below y=cot x contributes A2 = ∫_{π/4}^{π/2} cot x dx = [ln|sin x|]_{π/4}^{π/2} = ln √2 = (1/2)ln2. Total area = A1 + A2 = ln2.

Answer:

ln 2

Q.22Find the area of the region bounded by the parabola y = x^2 and the line y = −x + 2.v

Solution

Solve x^2 = −x + 2 ⇒ x^2 + x −2 = 0 ⇒ x = −2, 1. Area = ∫_{−2}^{1} [ (−x+2) − x^2 ] dx = [−x^2/2 + 2x − x^3/3]_{−2}^{1} = 9/2.

Answer:

9/2

Q.23A square field 0 ≤ x ≤ 4, 0 ≤ y ≤ 4 is to be divided along the curves y = x^2/4 and x = y^2/4 into three equal parts. Is it possible? If so, find the area for each person.v

Solution

Total area = 16. The region between the two parabolas is A = ∫_0^4 [2√x − x^2/4] dx = 32/3 − 16/3 = 16/3. By symmetry the two remaining regions are equal, so each of the three parts has area 16/3.

Answer:

Yes. Each area = 16/3

Q.24The curve y = (x – 2)² + 1 has a minimum point at P. A point Q on the curve is such that the slope of PQ is 2. Find the area bounded by the curve and the chord PQ.v

Solution

Answer:

x = 4 when x = 4, y = 5 Required Area = $\frac { 4 }{ 3 }$ sq. units

Q.25Find the area of the region common to the circle x^2 + y^2 = 16 and the parabola x = y^2/6 (equivalently y^2 = 6x).v

Solution

Intersections: substitute x = y^2/6 into circle ⇒ (y^2/6)^2 + y^2 = 16 ⇒ y^2 = 12 ⇒ y = ±2√3, x = 2. For 0 ≤ x ≤ 2 the points common satisfy |y| ≤ √(6x), while the circle gives larger |y|, so common area = ∫_0^2 2√(6x) dx = 2√6 * (2/3) x^{3/2}|_0^2 = (16√3)/3.

Answer:

16√3 / 3

EXERCISE 9.9EXERCISE 9.95 questions

Q.26Find, by integration, the volume generated by revolving about the x-axis the region enclosed by y = x^2, y = 0 and x = 1.v

Solution

Volume = π ∫_0^1 (x^2)^2 dx = π ∫_0^1 x^4 dx = π [x^5/5]_0^1 = π/5.

Answer:

π/5

Q.27Find, by integration, the volume generated by revolving about the x-axis the region enclosed by y = e^x − 2, y = 0, x = 0 and x = 1.v

Solution

Revolve about x-axis: V = π ∫_0^1 (e^x − 2)^2 dx = π ∫_0^1 (e^{2x} − 4e^x + 4) dx = π[ (1/2)e^{2x} − 4e^x + 4x ]_0^1 = π( (e^2/2 − 4e + 4) − (1/2 − 4) ) = π/2 (e^2 − 8e + 15 ).

Answer:

V = (π/2)(e^2 − 8e + 15)

Q.28Find, by integration, the volume generated by revolving about the y-axis the region enclosed by x = 1 + y^2 and y = 3 (and x = 0 as the inner boundary).v

Solution

Assume the region is 0 ≤ y ≤ 3, 0 ≤ x ≤ 1 + y^2. Revolving about the y-axis: V = π ∫_0^3 (1 + y^2)^2 dy = π ∫_0^3 (1 + 2y^2 + y^4) dy = π[ y + (2/3)y^3 + (1/5)y^5 ]_0^3 = π(3 + 18 + 243/5) = 348π/5.

Answer:

348π / 5

Q.29The region enclosed between the graphs of y = x and y = x^2 is denoted by R. Find the volume generated when R is rotated through 360° about the x-axis.v

Solution

Between x=0 and x=1 the upper curve is y=x and lower y=x^2. Revolve about x-axis: V = π ∫_0^1 (x^2 − x^4) dx = π[ x^3/3 − x^5/5 ]_0^1 = π(1/3 − 1/5) = 2π/15.

Answer:

2π/15

Q.30A watermelon is an ellipsoid formed by revolving an ellipse with major axis 20 cm and minor axis 10 cm about its major axis. Find its volume.v

Solution

Semi-axes: a = 10 (major/2), b = 5 (minor/2). Volume of revolution (ellipsoid) = (4/3)π a b^2 = (4/3)π ·10·5^2 = 1000π/3 cm^3.

Answer:

1000π / 3 cm^3

Choose the correctChoose the correct20 questions

Q.31The value of $\int_{0}^{2/3}$ $\frac { dx }{ \sqrt{4-9x^2} }$
(A) $\frac { π }{ 6 }$ (B) $\frac { π }{ 2 }$ (C) $\frac { π }{ 4 }$ (D) πv

Solution

(a) $\frac { π }{ 6 }$
Hint:

Answer:

$\frac { π }{ 6 }$

Q.32The value of $\int_{-1}^{2}$ |x| dx
(A) $\frac { 1 }{ 2 }$ (B) $\frac { 3 }{ 2 }$ (C) $\frac { 5 }{ 2 }$ (D) $\frac { 7 }{ 2 }$v

Solution

Answer:

$\frac { 5 }{ 2 }$

Q.33For any value of n ∈ Z $\int_{0}^{π}$ e cos²x cos3 [(2n + 1) x] dx
(A) $\frac { π }{ 2 }$ (B) π (C) 0 (D) 2v

Solution

Answer:

Q.34The value of $\int_{-π/2}^{π/2}$ sin² x cos x dx
(A) $\frac { 3 }{ 2 }$ (B) $\frac { 1 }{ 2 }$ (C) 0 (D) $\frac { 2 }{ 3 }$v

Solution

(d) $\frac { 2 }{ 3 }$
Hint:
f(x) = sin²x cos x
f(-x) = sin²(-x) cos(-x) = sin²x cos x
f(x) = f(-x)
f is an even function

Answer:

$\frac { 2 }{ 3 }$

Q.35The value of $int_{-4}^{4}$ $left[ an^{-1}rac { x^2 }{ x^4+1 }+ an^{-1}rac { x^4+1 }{ x^2 } ight]$ dx is
(A) $pi$ (B) $2pi$ (C) $3pi$ (D) $4pi$v

Solution

(d) $4pi$
Hint:
For $x e0$, $rac { x^2 }{ x^4+1 }>0$, so using $ an^{-1}u+ an^{-1}rac { 1 }{ u }=rac { pi }{ 2 }$ with $u=rac { x^2 }{ x^4+1 }$, the integrand $=rac { pi }{ 2 }$.
$int_{-4}^{4}rac { pi }{ 2 }$ dx $=rac { pi }{ 2 } imes8=4pi$

Answer:

$4pi$

Q.36The value of $\int_{-π/4}^{π/4}$ ($\frac { 2x^7-3x^5+7x^3-x+1 }{ cos^2x }$) dx is
(A) 4 (B) 3 (C) 2 (D) 0v

Solution

(c) 2
Hint:

Odd function – 3 Odd function + 7 Odd function – Odd function + even function
= 0 + 2 $\int_{0}^{π/4}$ sec² x dx
= 2 [tan x]$_{0}^{π/4}$
= 2(1 – 0)
= 2

Answer:

Q.7If f(x) = $\int_{0}^{x}$ cos t dt, then $\frac { df }{ dx }$ =
(A) cos x – x sin x (B) sin x + x cos x (C) x cos x (D) x sin xv

Solution

(c) x cos x
Hint:
f(x) = $\int_{0}^{x}$ cos t dt
u = t, dv = cos t dt
u’ = 1, v = sin t
v 1 = -cos t
f(x) = [t sin t + cos t]$_{0}^{x}$
f(x) = x sin x + cos x – 1
$\frac { df }{ dx }$ = x cos + sin x – sin x
$\frac { df }{ dx }$ = x cos x

Answer:

x cos x

Q.37The area between y² = 4x and its latus rectum is
(A) $\frac { 2 }{ 3 }$ (B) $\frac { 4 }{ 3 }$ (C) $\frac { 8 }{ 3 }$ (D) $\frac { 5 }{ 3 }$v

Solution

Answer:

$\frac { 8 }{ 3 }$

Q.38The value of $\int_{0}^{1}$ x (1 – x) 99 dx is
(A) $\frac { 1 }{ 11000 }$ (B) $\frac { 1 }{ 10100}$ (C) $\frac { 1 }{ 10010 }$ (D) $\frac { 1 }{ 10001 }$v

Solution

(b) $\frac { 1 }{ 10100}$
Hint:

Answer:

$\frac { 1 }{ 10100}$

Q.39The value of $\int_{0}^{1}$ x (1 – x) 99 dx is
(A) $\frac { 1 }{ 11000 }$ (B) $\frac { 1 }{ 10100}$ (C) $\frac { 1 }{ 10010 }$ (D) $\frac { 1 }{ 10001 }$v

Solution

(b) $\frac { 1 }{ 10100}$
Hint:

Answer:

$\frac { 1 }{ 10100}$

Q.40If $\frac { Γ(n+2) }{ Γn }$ = 90 then n is
(A) 10 (B) 5 (C) 8 (D) 9v

Solution

(d) 9
Hint:

n² + n = 90
n² + n – 90 = 0
(n + 10) (n – 9) = 0
n = 9

Answer:

Q.41The value of $\int_{0}^{π/6}$ cos³ 3x dx is
(A) $\frac { 2 }{ 3 }$ (B) $\frac { 2 }{ 9}$ (C) $\frac { 1 }{ 9 }$ (D) $\frac { 1 }{ 3 }$v

Solution

(b) $\frac { 1 }{ 10100}$
Hint:
$\int_{0}^{π/6}$ cos³ 3xdx = $\frac { 1 }{ 3 }$ × $\frac { 2 }{ 3 }$ × 1
= $\frac { 2 }{ 9 }$

Answer:

$\frac { 2 }{ 9}$

Q.42The value of $\int_{0}^{π/6}$ cos³ 3x dx is
(A) $\frac { 2 }{ 3 }$ (B) $\frac { 2 }{ 9}$ (C) $\frac { 1 }{ 9 }$ (D) $\frac { 1 }{ 3 }$v

Solution

(b) $\frac { 1 }{ 10100}$
Hint:
$\int_{0}^{π/6}$ cos³ 3xdx = $\frac { 1 }{ 3 }$ × $\frac { 2 }{ 3 }$ × 1
= $\frac { 2 }{ 9 }$

Answer:

$\frac { 2 }{ 9}$

Q.43The value of $\int_{0}^{π}$ sin 4 x dx
(A) $\frac { 3π }{ 10 }$ (B) $\frac { 3π }{ 8}$ (C) $\frac { 3π }{ 4 }$ (D) $\frac { 3π }{ 2 }$v

Solution

(b) $\frac { 3π }{ 8}$
Hint:
(x) = sin 4 x dx
f(π – x) = sin 4 (π – x) = sin 4 x

Answer:

$\frac { 3π }{ 8}$

Q.44The value of $\int_{0}^{∞}$ e -3x x² dx
(A) $\frac { 7 }{ 27 }$ (B) $\frac { 5 }{ 27 }$ (C) $\frac { 4 }{ 27 }$ (D) $\frac { 2 }{ 27 }$v

Solution

(d) $\frac { 2 }{ 27 }$
Hint:

Answer:

$\frac { 2 }{ 27 }$

Q.45The volume of solid of revolution of the region bounded by y² = x(a – x) about the x-axis is
(A) πa³ (B) $\frac { πa^3 }{ 4 }$ (C) $\frac { πa^3 }{ 5 }$ (D) $\frac { πa^3 }{ 6 }$v

Solution

(d) $\frac { πa^3 }{ 6 }$
Hint:
y² = x(a – x)
To find limit put y = 0
x (a – x) = 0
x = 0, x = a

Answer:

$\frac { πa^3 }{ 6 }$

Q.46If f(x) = $\int_{1}^{x}$ $\frac { e^{sinx} }{ u }$ du, x > 1 and $\int_{1}^{3}$ $\frac { e^{sinx^2} }{ x }$ dx = $\frac { πa^3 }{ 6 }$ [f(a) – f(1)] then one of the possible value of a is
(A) 3 (B) 6 (C) 9 (D) 5v

Solution

Answer:

Q.47The value of $\int_{0}^{1}$ (sin -1 x)² dx is
(A) $\frac { π^2 }{ 4 }$ – 1 (B) $\frac { π^2 }{ 4 }$ + 2 (C) $\frac { π^2 }{ 4 }$ + 1 (D) $\frac { π^2 }{ 4 }$ – 2v

Solution

(d) $\frac { π^2 }{ 4 }$ – 2
Hint:

Answer:

$\frac { π^2 }{ 4 }$ – 2

Q.48The value of $\int_{0}^{1}$ (sin -1 x)² dx is
(A) $\frac { π^2 }{ 4 }$ – 1 (B) $\frac { π^2 }{ 4 }$ + 2 (C) $\frac { π^2 }{ 4 }$ + 1 (D) $\frac { π^2 }{ 4 }$ – 2v

Solution

(d) $\frac { π^2 }{ 4 }$ – 2
Hint:

Answer:

$\frac { π^2 }{ 4 }$ – 2

Q.49If $\int_{0}^{x}$ f(t) dt = x + $\int_{x}^{1}$ f(t) dt, then the value of f(1) is
(A) $\frac { 1 }{ 2 }$ (B) 2 (C) 1 (D) $\frac { 3 }{ 4 }$v

Solution

(a) $\frac { 1 }{ 2 }$
Hint:

= 1 + 0 – x f(x)
f(x) + x f(x) = 1
f(x)[1 + x] = 1
when x = 1,
f(1)(2) = 1
f(1) = $\frac { 1 }{ 2 }$

Answer:

$\frac { 1 }{ 2 }$

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