Definite Integration

Q1 FREE PREVIEW

PYQ

\(\int_a^b f(x) d x\) is equal to :

a

\(\int_a^b f(a-x) d x\)

b

\(\int_a^b f(a+b-x) d x\)

c

\(\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{x}-(\mathrm{a}+\mathrm{b})) \mathrm{dx}\)

d

\(\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}((\mathrm{a}-\mathrm{x})+(\mathrm{b}-\mathrm{x})) \mathrm{dx}\)

✓ Correct answer: b)

\(\int_a^b f(a+b-x) d x\)

Explanation

Let \(x=a+b-u\).

\(dx=-du.\)

When \(x=a, u=b\);

when \(x=b, u=a\).

\({\int }_{a}^{b}f\left(x\right)dx={\int }_{u=b}^{u=a}f\left(a+b-u\right)\left(-du\right)\)

\(\Rightarrow {\int }_{a}^{b}f\left(x\right)dx={\int }_{u=a}^{u=b}f\left(a+b-u\right)du\)

Rename \(u\) to \(x\) :

\(\int_a^b f(x) d x=\int_a^b f(a+b-x) d x\).

Q2

PYQ

The value of \({\int }_{-1}^{1}x|x|\mathrm{d}x\) is :

a

\(\frac{1}{6}\)

b

\(\frac{1}{3}\)

c

\(-\frac{1}{6}\)

d

\(0\)

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Q3

PYQ

\(\int_0^{\pi / 2} \frac{\sin x-\cos x}{1+\sin x \cos x} d x\) is equal to :

a

\(\pi\)

b

Zero (0)

c

\(\int_0^{\pi / 2} \frac{2 \sin x}{1+\sin x \cos x} d x\)

d

\(\frac{\pi^2}{4}\)

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Q4

PYQ

\(\int_0^{\pi / 2} \frac{\sin x-\cos x}{1+\sin x \cos x} d x\) is equal to :

a

\(\pi\)

b

Zero (0)

c

\(\int_0^{\pi / 2} \frac{2 \sin x}{1+\sin x \cos x} d x\)

d

\(\frac{\pi^2}{4}\)

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Q5

PYQ

The value of \({\int }_{-1}^{1}x|x|\mathrm{d}x\) is :

a

\(\frac{1}{6}\)

b

\(\frac{1}{3}\)

c

\(-\frac{1}{6}\)

d

\(0\)

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Q6

PYQ

\(\int_a^b f(x) d x\) is equal to :

a

\(\int_a^b f(a-x) d x\)

b

\(\int_a^b f(a+b-x) d x\)

c

\(\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{x}-(\mathrm{a}+\mathrm{b})) \mathrm{dx}\)

d

\(\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}((\mathrm{a}-\mathrm{x})+(\mathrm{b}-\mathrm{x})) \mathrm{dx}\)

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Q7

PYQ

Let \(f\left(x\right)={[x]}^{2}−\left[x+3\right]−3,\text{ }x\in R,\) where [.] is the greatest integer function. Then

[JEE Main 2026, 22 Jan (Shift 2)]

a

\(f(x)=0\) for finitely many values of x

b

\(f(x)<0\) only for \(x\in [−1,\text{ }3)\)

c

\(f\left(x\right)=0\) only for \(x\in [4,\text{ }∞)\)

d

\({\int }_{0}^{2}f(x)dx=−6\)

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Q8

PYQ

The value of the integral \({\int }_{\frac{\pi }{24}}^{\frac{5\pi }{24}}\frac{\mathrm{d}x}{1+\sqrt[3]{\tan 2x}}\) is:

[JEE Main 2026, 23 Jan (Shift 1)]

a

\(\frac{\pi }{18}\)

b

\(\frac{\pi }{3}\)

c

\(\frac{\pi }{6}\)

d

\(\frac{\pi }{12}\)

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Q9

PYQ

The integral \({\int }_{1/4}^{3/4}\cos \left(2co{t}^{-1}\sqrt{\frac{1-x}{1+x}}\right)dx\) is equal to:

[JEE Main 2024, 09 Apr (Shift 2)]

a

-1/2

b

1/4

c

1/2

d

-1/4

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Q10

PYQ

The integral \({\int }_{1/4}^{3/4}\cos \left(2co{t}^{-1}\sqrt{\frac{1-x}{1+x}}\right)dx\) is equal to:

[JEE Main 2024, 09 Apr (Shift 2)]

a

-1/2

b

1/4

c

1/2

d

-1/4

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Q11

PYQ

If \(f(2a-x)=f(x)\) , then\({\int }_{0}^{2a}f(x)dx\) is

a

\({\int }_{0}^{2a}f\left(\frac{x}{2}\right)dx\)

b

\({\int }_{0}^{a}f(x)dx\)

c

\(2{\int }_{a}^{0}f(x)dx\)

d

\(2{\int }_{0}^{a}f(x)dx\)

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Q12

PYQ

If \(f(2a-x)=f(x)\) , then\({\int }_{0}^{2a}f\left(x\right)dx\) is

a

\({\int }_{0}^{2a}f\left(\frac{x}{2}\right)dx\)

b

\({\int }_{0}^{a}f(x)dx\)

c

\(2{\int }_{a}^{0}f(x)dx\)

d

\(2{\int }_{0}^{a}f(x)dx\)

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