2nd PUC Maths Question Bank Chapter 7 Integrals

Students can Download 2nd PUC Maths Chapter 7 Integrals Questions and Answers, Notes Pdf, 2nd PUC Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

Karnataka 2nd PUC Maths Question Bank Chapter 7 Integrals

2nd PUC Maths Integrals One Mark Questions and Answers

Question 1.
Evaluate ∫(1 – x) √x dx.
Answer:

Question 2.
Write the integral of \(\frac{1}{x \sqrt{x^{2}-1}}\),x > 1 with respect to x.
Answer:

Question 3.
Write the antiderivative of e^2x with respect to x.
Answer:
∫e^2x dx = \(\frac{e^{2 x}}{2}\) + C

Question 4.
Sin 2x.
Answer:
∫Sin 2x dx = \(-\frac{1}{2}\)cos 2x + C

Question 5.
Cos 3x
Answer:
∫cos 3x dx = \(\frac { 1 }{ 3 }\)sin 3x + C

Question 6.
(ax + b)².
Answer:
∫(ax + b)² dx = \(\frac{1}{3 a}\)(ax + b)³ + C

Question 7.
sin 2x – 4e^3x.
Answer:
sin 2x – 4e^3x dx

Question 8.
∫(4e^3x + 1) dx.
Answer:

Question 9.

Answer:

Question 10.
∫(ax² + bx + c) dx.
Answer:
∫(ax² + bx + c) dx = ∫ax² dx + ∫bx dx + c∫1 dx

Question 11.
∫(2x² + e^x) dx.
Answer:
∫(2x² + e^x) dx = 2∫x²dx + ∫e^x dx = \(\frac{2 x^{3}}{3}\) + e^x + C

Question 12.

Answer:

2nd PUC Maths Integrals Two Marks/ Three Marks Questions and Answers

Question 1.

Answer:

Question 2.
∫log x dx.
Answer:
∫log x dx = ∫log x. 1 dx = log x. x – ∫x. \(\frac{1}{x}\) dx
= x log x – ∫1 dx = x log x – x + c.

Question 3.
∫log(sin x).(cot x) dx.
Answer:

Question 4.

Answer:

Question 5.
∫e^x sec x(1 + tan x)dx.
Answer:
∫e^x sec x(1 + tan x)dx
= ∫e^x(sec x + sec x tan x)dx.
= e^x sec x + c
∵ ∫e^x (f(x) + f'(x))dx = e^xf(x).

Question 6.

Answer:

Question 7.

Answer:

Question 8.

Answer:

Question 9.
∫(1 + x) √x dx.
Answer:

Question 10.
∫√x(3x² + 2x + 3) dx.
Answer:

Question 11.
∫(2x – 3cosx + e^x) dx.
Answer:
= x² – 3 sin x + e^x + C.

Question 12.
∫(2x² – 3 sin x + 5√x)dx.
Answer:

Question 13.
∫sec(sec x + tan x) dx.
Answer:
∫sec(sec x + tan x)dx = ∫(sec² x + sec x.tan x)
= ∫sec² x dx + ∫sec x.tan x dx = tan x + sec x + C.

Question 14.

Answer:

Question 15.

Answer:

Question 16.

Answer:

Question 17.

Answer:

Question 18.

Answer:

Question 19.
∫sin x sin(cos x) dx.
Answer:

Question 20.

Answer:

Question 21.
∫x\(\sqrt{x+2}\) dx.
Answer:

Question 22.
∫\(x \sqrt{1+2 x^{2}} d x\) dx.
Answer:

Question 23.
∫(4x + 2) \(\sqrt{x^{2}+x+1}\) dx.
Answer:

Question 24.

Answer:

Question 25.

Answer:

Question 26.
∫(x³ – 1)^1/3 x⁵ dx.
Answer:

Question 27.

Answer:

Question 28.

Answer:

Question 29.

Answer:

Question 30.
∫e^(2x+3) dx.
Answer:

Question 31.

Answer:

Question 32.

Answer:

Question 33.

Answer:

Question 34.

Answer:

Question 35.
∫tan² (2x – 30) dx.
Answer:

Question 36.
∫sec²(7 – 4x) dx.
Answer:

Question 37.

Answer:

Question 38.

Answer:

Question 39.

Answer:

Question 40.

Answer:

Question 41.
∫\(\sqrt{\sin 2 x} \cos 2 x d x\)
Answer:

Question 42.

Answer:

Question 43.

Answer:

Question 44.

Answer:

Question 45.
∫x sin x dx.
Answer:
∫x sin x dx.
On taking x as first function and sin x as second function and integrating by parts, we get

= -x cos x + ∫1.cos x dx
⇒ I = -x cosx + Sinx + C.

Question 46.
∫x sin 3x dx.
Answer:
Let I = ∫x sin 3x dx
On taking x as first function and sin 3x as second function and integrating by parts, we get

Question 47.
∫x log x dx.
Answer:
Let I = ∫x log x dx
On taking log x as first function and x as second function and integrating by parts, we get
(∵ log function comes before algebraic function in ILATE)

Question 48.
∫x log 2x dx.
Answer:
Let I = ∫x log 2x dx
On taking log 2x as first function and x as second function and integrating by parts, we get
(∵ log function cornes before algebraic function in ILATE)

Question 49.
∫x² log x dx.
Answer:
On taking log x as first function and X2 as second function and integrating by parts, we get
(∵ log function comes before algebraic function in ILATE)

Question 50.
∫\(\sqrt{4-x^{2}} d x\)
Answer:

Question 51.
∫\(\sqrt{1-4 x^{2}} d x\)
Answer:

Question 52.
∫\(\sqrt{x^{2}+4 x+6} d x\)
Answer:

Question 53.
∫\(\sqrt{x^{2}+4 x+1} d x\)
Answer:

Question 54.
∫\(\sqrt{1-4 x-x^{2}} d x\)
Answer:

Question 55.
∫\(\sqrt{x^{2}+4 x-5} d x\)
Answer:

Question 56.

Answer:

Question 57.

Answer:

Question 58.

Answer:

Question 59.

Answer:

Question 60.

Answer:

2nd PUC Maths Integrals Five Marks Questions and Answers

Question 1.
Find the integral of \(\frac{1}{\sqrt{a^{2}-x^{2}}}\) with respect to x and hence evaluate \(\int \frac{d x}{\sqrt{7-6 x-x^{2}}}\)
Answer:

Question 2.
Find the integral of \(\frac{1}{x^{2}+a^{2}}\) r with respect to x and hence evaluate \(\int \frac{1}{x^{2}+2 x+2} d x\)
Answer:

Question 3.
Find the integral of \(\frac{1}{x^{2}-a^{2}}\) w.r. t. x and hence evaluate \(\int \frac{d x}{3 x^{2}+13 x-10}\)
Answer:

Question 4.
Find the integral of \(\frac{1}{a^{2}-x^{2}}\) w.r. t. x and hence evaluate \(\int \frac{d x}{16-x^{2}}\)
Answer:

Question 5.
Find the integral of \(\frac{1}{a^{2}+x^{2}}\) w.r. t. x and hence evaluate \(\int \frac{d x}{x^{2}-16 x+13}\)
Answer:

Question 6.
Find the integral of \(\frac{1}{\sqrt{x^{2}+a^{2}}}\) w.r. t. x and hence evaluate \(\int \frac{d x}{\sqrt{x^{2}+7}}\) June 2014
Answer:

Question 7.
Find the integral of \(\frac{1}{\sqrt{x^{2}-a^{2}}}\) w.r. t. x and hence evaluate \(\int \frac{d x}{\sqrt{x^{2}+6 x-7}}\). June 2014.
Answer:

Question 8.
Find the integral of \(\frac{1}{\sqrt{a^{2}-x^{2}}}\) w.r. t. x and hence evaluate \(\int \frac{d x}{\sqrt{7-6 x-x^{2}}}\) June 2014.
Answer:

Question 9.
Find the integral of \(\sqrt{x^{2}+a^{2}}\) w.r. t. x and hence evaluate \(\int \sqrt{4 x^{2}+9} d x\)
Answer:
Let I = ∫\(\sqrt{x^{2}+a^{2}}\)
Taking constant function I as the second function and integrating by parts, we have

Question 10.
Find the integral of \(\frac{1}{x^{2}+a^{2}}\) w.r. t. x and hence evaluate \(\int \frac{d x}{x^{2}+2 x+2}\)
Answer:

Question 11.
Find the integral of \(\sqrt{x^{2}-a^{2}}\) w.r. t. x and hence evaluate \(\int \sqrt{x^{2}+4 x-5} d x\)
Answer:

Question 12.
Find the integral of \(\sqrt{a^{2}-x^{2}}\) w.r. t. x and hence evaluate \(\int \sqrt{3-2 x-x^{2}} d x\)
Answer:


3 – 2x – x² = – [x² + 2x – 3]
= – [x² + 2x + 1 – 1 – 3]
= – [(x + 1)² – 4]
= 4 – (x + 1)²

Question 13.
Find the integral of [ latex]\sqrt{x^{2}+a^{2}}[/latex] w.r. t. x and hence evaluate \(\int \sqrt{x^{2}+4 x+6} d x\) (June 2015)
Answer:

Question 14.
Find the integral of \(\sqrt{x^{2}+a^{2}}\) w.r. t. x and hence evaluate \(\int \sqrt{4 x^{2}+9} d x\)
Answer:
Let I = ∫\(\sqrt{x^{2}+a^{2}}\)
Taking cbnstant function I as the second function and integrating by parts, we have

2nd PUC Maths Integrals Six Marks Questions and Answers

Question 1.

Answer:

Question 2.

Answer:

Question 3.

Answer:

t = -x in the first integral on the right hand side.
dt =-dt When x = -a, t = a andwhen
x = 0, t = 0 Also x = -t.

f(-x) = sin⁵ (-x)cos⁴ (-x) = -sin⁵ x cos⁴ x = -f(x) i.e.f is an odd function
Therefore, by P₇ (ii), I = 0

Question 4.

Answer:

Question 5.
Prove that \(\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\) and hence evaluate the following:
Answer:
\(\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\)
Proof: Put u = a – x on R. H. S. Then du = – dx
When x = 0, u = a and when x = a, u = 0

Answer:

Answer:

Answer:

Answer”

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