The area of the closed igure bounded by $x = - 1,\;x = 2$ and $y = \left\{ {\begin{array}{*{20}{c}} { - {x^2} + 2\;,\;x \le 1}\\ {2x - 1\;\;,\;\;x > 1} \end{array}} \right.$ and the $x$-axis is

$\frac{{16}}{3}\,{\rm{sq unit}}$

$\frac{{10}}{3}\,{\rm{sq unit}}$

$\frac{{13}}{3}\,{\rm{sq unit}}$

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JEE • MATHEMATICS • DEFINITE INTEGRATION

Definite integration practice · Easy

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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391 Easy

391 of 791

Application +4 | -1

\int \nolimits_0^a \,\,\sqrt {{a^2} - \,\,{x^2}\,} \;dx

is equal to

View answer and explanation

Correct option: D

\frac{1}{4}\,\,\pi {a^2}

The correct answer is (1/4)πa² because the integral represents the area of a quarter circle with radius a.

392 Easy

392 of 791

Application +4 | -1

\lim\limits_{n \to \infty } \,\,{\left( {\frac{1}{n}\,\, + \,\,\frac{1}{{n + 1}} + \,\,... + \,\,\frac{1}{{3\,n}}} \right)^{}}

is equal to

View answer and explanation

Correct option: B

\log \,\,3

Option B is correct because the limit of the sum can be expressed as a definite integral, which evaluates to log 3.

393 Easy

393 of 791

Analyzing +4 | -1

Let

f:R \to R

be a continuous function satisfying

f\left( x \right) + f\left( {x + k} \right) = n,

for all

x \in R

where

k > 0

and

n

is a positive integer. If

{l_1} = \int\limits_0^{4nk} {f\left( x \right)} dx

and

{l_2} = \int\limits_{ - k}^{3k} {f\left( x \right)} dx,

then

View answer and explanation

Correct option: C

{l_1} + n{l_2} = 4{n^2}k

Option D (

{l_1} + n{l_2} = 4{n^2}k

) is correct because it is derived from the periodicity of the function and the properties of definite integrals.

394 Easy

394 of 791

Application +4 | -1

Area bounded by the curve

y=x e^{x^2}, x

-axis and the ordinates

x=0, x=a

View answer and explanation

Correct option: B

\frac{e^{a^2}-1}{2}

sq. unit

Option B (

\frac{{e^{a^2}-1}}{{2}}

sq. unit) is correct because it represents the definite integral of the given function from 0 to a, which calculates the area under the curve.

395 Easy

395 of 791

Analyzing +4 | -1

I = \int \limits_{ - \pi }^\pi \,\,\frac{{{{\sin }^2}\,\,x}}{{1 + {a^x}}}\,\,dx,\,\,a > 0

, then

I

equals

View answer and explanation

Correct option: B

\frac{\pi }{2}

Option B is correct because by using the property of definite integrals, the integral simplifies to π/2.

396 Easy

396 of 791

Application +4 | -1

The area of the closed igure bounded by

x = - 1,\;x = 2

and

y = \left\{ {\begin{array}{cccccccccccccccccccc} { - {x^2} + 2\;,\;x \le 1}\\ {2x - 1\;\;,\;\;x > 1} \end{array}} \right.

and the

x

-axis is

View answer and explanation

Correct option: A

\frac{{16}}{3}\,\mathrm{sq unit}

Option A (

\frac{{16}}{3}\,\mathrm{sq unit}

) is correct because it represents the definite integral of the piecewise function over the given interval.

397 Easy

397 of 791

Application +4 | -1

\int\limits_{{-a}}^a \sqrt{\frac{a-x}{a+x}} d x=k \pi

, then

k=

View answer and explanation

Correct option: D

a

Option D (

a

) is correct because the integral evaluates to

a\pi

, and comparing this with

k\pi

gives

k=a

398 Easy

398 of 791

Application +4 | -1

f(y)=e^y, g(y)=y ; y > 0

and

F(t)=\int\limits_{0}^t f(t-y) g(y) d y

, then

View answer and explanation

Correct option: C

F(t)=e^t-(1+t)

Option A (

\frac{3}{4}(2)^{4 / 3}-\frac{3}{4}

) is correct because the limit of the sum is converted to a definite integral.

399 Easy

399 of 791

Application +4 | -1

Let

f: R \rightarrow R

be a continuous function which satisfies

f(2 x)=3 f(x) \,\forall \,x \in R

. If

\int\limits_{0}^1 f(x) d x=1

, then find the value of

\frac{1}{2} \int\limits_{1}^2 f(x) d x

View answer and explanation

Correct option: A

2.5

Option A (2.5) is correct because using the given functional equation and integral property, we can find the value of the required integral.

400 Easy

400 of 791

Application +4 | -1

\int\limits_{0}^2\left|x^2+2 x-3\right| \,d x

is equal to

View answer and explanation

Correct option: A

Option A (4) is correct because the integral of the absolute value of the quadratic function is calculated by splitting the integral at the root within the interval.

401 Easy

401 of 791

Application +4 | -1

\int \nolimits_0^1 \,\,\frac{1}{{x + \sqrt x }}\,\,dx

is equal to

View answer and explanation

Correct option: C

\log \,\,4

Option C is correct because the integral evaluates to

\log 4

after a suitable substitution.

402 Easy

402 of 791

Application +4 | -1

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {{e^{\cos x}}\,\,\sin \,\,x,}\\ {2,} \end{array}\,\,\begin{array}{cccccccccccccccccccc} {\left| {\,x\,} \right|\,\, \le \,\,2}\\ {\mathrm{otherwise}} \end{array},} \right.

then

\int \nolimits_{ - 2}^3 \,\,f\left( x \right)\,\,dx

is equal to

View answer and explanation

Correct option: C

\mathrm{2}

Option C (2) is correct because the integral can be split into two parts, where the first part is zero due to the odd function and the second part evaluates to 2.

403 Easy

403 of 791

Application +4 | -1

The area of the region (in sq. units), in the first quadrant bounded by the parabola

y=9 x^2

and the lines

x=0, y=1

and

y=4

, is:

View answer and explanation

Correct option: D

14 / 9

Option D (14 / 9) is correct because the area under the curve

y=9x^2

between y=1 and y=4 is 14/9.

404 Easy

404 of 791

Application +4 | -1

The value of

\int\limits_0^\pi {\frac{{{e^{\cos x}}\sin x}}{{(1 + {{\cos }^2}x)({e^{\cos x}} + {e^{ - \cos x}})}}dx}

is equal to:

View answer and explanation

Correct option: C

\frac{\pi }{4}\,

Option C (π/4) is correct because the integral can be simplified using substitution and properties of definite integrals.

405 Easy

405 of 791

Application +4 | -1

Area under the curve

y=x^2-4 x

within the

x

-axis and the line

x=2

, is

View answer and explanation

Correct option: A

\frac{16}{3}

sq.unit

Option A (

\frac{16}{3}

sq.unit) is correct because the definite integral of

x^2 - 4x

from 0 to 2 gives the area under the curve, considering the function is negative in this interval, we take the absolute value.

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SEQUENCE OF SERIES 1764 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs DIFFERENTIAL EQUATIONS 977 MCQs

406 Easy

406 of 791

Analyzing +4 | -1

\int\limits_{{\pi / 5}}^{3 \pi / 10} \frac{\sin x}{\sin x+\cos x} d x

is equal to

View answer and explanation

Correct option: D

None of these

The correct answer is None of these because the integral evaluates to

\frac{{\pi}}{{20}}

, which is not among the given options.

407 Easy

407 of 791

Application +4 | -1

f(x)

and

g(x)

are two differentiable functions on

[0,2]

such that

f^{\prime \prime}(x)-g^{\prime \prime}(x)=0, f^{\prime}(1)=2 g^{\prime}(1)=4 f(2)=3 g(2)=9

then

f(x)-g(x)

x=3 / 2

View answer and explanation

Correct option: D

Option D (5) is correct because by integrating twice and using the given conditions, we can find the value of f(x) - g(x) at x=3/2.

408 Easy

408 of 791

Application +4 | -1

\int \nolimits_0^{\pi /2} \,\,\frac{{\cos \,\,x}}{{1\,\, + \,\,\sin \,\,x}}\,\,\;dx

is equal to

View answer and explanation

Correct option: A

\log \,\,2

Option A (

\log \,2

) is correct because the definite integral of

\frac{{\cos \,x}}{{1 + \sin \,x}}

from 0 to

\frac{\pi}{2}

\log 2

409 Easy

409 of 791

Analyzing +4 | -1

The value of

\int \nolimits_0^{\pi /2} \,\,\frac{{\cos \,\,3x\,\, + \,\,1}}{{2\,\,\cos \,\,x\,\, - \,\,1}}\,\,dx

View answer and explanation

Correct option: B

\mathrm{1}

Option B (1) is correct because after applying trigonometric identities and simplifying, the integral evaluates to 1.

410 Easy

410 of 791

Application +4 | -1

\int \limits_0^{{\pi ^2}/4} \sin \,\,\sqrt x \,dx

is equals to

View answer and explanation

Correct option: C

\mathrm{2}

Option C,

\mathrm{2}

, is correct because after substitution and integration by parts, the definite integral evaluates to 2.

411 Easy

411 of 791

Application +4 | -1

Area enclosed by the parabola

a y=3\left(a^2-x^2\right)

and

x

-axis is

View answer and explanation

Correct option: A

4 a^2 \mathrm{sq}

, unit

Option A (

4 a^2

sq, unit) is correct because it represents the area enclosed by the parabola and the x-axis, calculated by integrating the function between its roots.

412 Easy

412 of 791

Application +4 | -1

\int \nolimits_0^{\pi /2} \frac{{2\,\sqrt {\cos \,\,\theta } }}{{3\,(\sqrt {\sin \,\,\theta } \,\, + \,\,\sqrt {\cos \,\,\theta )} }}\,\,d\mathrm{\theta }

is equal to

View answer and explanation

Correct option: A

\frac{\pi }{6}

Option A (

\frac{\pi}{6}

) is correct because using the property of definite integrals and adding the two forms of the integral simplifies the expression, leading to the result.

413 Easy

413 of 791

Application +4 | -1

\int \nolimits_0^1 \,\,\frac{{dx}}{{x + \sqrt {1 - {x^2}} }}

is equal to

View answer and explanation

Correct option: D

\pi /4

Option D (

\pi /4

) is correct because the integral can be solved using trigonometric substitution and properties of definite integrals.

414 Easy

414 of 791

Application +4 | -1

The value of the integral

\int\limits_{{\pi / 6}}^{\pi / 3} \frac{1}{1+\sqrt{\tan x}} d x,

View answer and explanation

Correct option: C

\frac{\pi}{12}

Option C is correct because by using the property

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

, the integral simplifies to

\frac{b-a}{2}

415 Easy

415 of 791

Analyzing +4 | -1

The value of

\int_0^{2\pi } {\frac{{x{{\sin }^8}x}}{{{{\sin }^8}x + {{\cos }^8}x}}dx}

is equal to

View answer and explanation

Correct option: D

{\pi ^2}

Option D (

{\pi ^2}

) is correct because it is the value of the definite integral after applying properties of definite integrals to simplify it.

416 Easy

416 of 791

Application +4 | -1

Area bounded by the curve

x y-3 x-2 y-10=0, x

-axis and the lines

x=3, x=4

View answer and explanation

Correct option: C

16 \log 2+3

Option C (

16 \log 2+3

) is correct because it represents the definite integral of the given function between the specified limits, calculated after proper simplification.

417 Easy

417 of 791

Application +4 | -1

The value of integral

\int \nolimits_0^4 \,\,\left| {\,x\,\, - \,\,1\,} \right|\,\,dx\;\,\,\mathrm{is}

View answer and explanation

Correct option: B

\mathrm{5}

Option B (

\mathrm{5}

) is correct because the integral of the absolute value function is evaluated by splitting the integral at the point where the function changes sign.

418 Easy

418 of 791

Analyzing +4 | -1

The value of

\int\limits_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{(2+3 \sin x)}{\sin x(1+\cos x)} d x

is equal to

View answer and explanation

Correct option: D

\frac{10}{3}-\sqrt{3}+\log _{e} \sqrt{3}

Option D (

\frac{10}{3}-\sqrt{3}+\log _{e} \sqrt{3}

) is correct because the integral is solved by partial fraction decomposition and trigonometric substitution.

419 Easy

419 of 791

Analyzing +4 | -1

Let

a, b, c

be non-zero real numbers such that

\int\limits_{{0}}^{1}\left(1+\cos ^8 x\right)\left(a x^2+b x+c\right) d x= \int\limits_{0}^2\left(1+\cos ^8 x\right) \times\left(a x^2+b x+c\right) d x

Then the quadratic equation

a x^2+b x+c=0

has

View answer and explanation

Correct option: B

at least one root in

(0,2)

Option B (at least one root in

(0,2)

) is correct because by applying Rolle's theorem to the integral, we can conclude that there exists at least one root in the interval (1,2), which is within (0,2).

420 Easy

420 of 791

Application +4 | -1

The area, enclosed by the curves

y = \sin x + \cos x\mathrm{ }\,\,\mathrm{and}\,\,\mathrm{ }y = |\cos x - \sin x|

and the lines

x = 0,\,\,x = \frac{\pi }{2},

View answer and explanation

Correct option: A

2\sqrt 2 (\sqrt 2 - 1)

Option A (

2\sqrt 2 (\sqrt 2 - 1)

) is correct because it represents the area between the curves, calculated by integrating the difference of the functions over the given interval.

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Review DEFINITE INTEGRATION in your textbook or class notes, then use this set to check your understanding.

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