NIMCET 2025 Paper

Section: Mathematics

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Question : 13 of 120

Marks: +1, -0

The value of

{\;{d}/{d x} ∫↙{ sin \;^2 x}↖{2 sin \;x} e^{t^2} \dt }

at

x=π

is:

-1
1
2
-2

Solution:

Here:
- Upper limit

b(x)=2 sin \;x

- Lower limit

a(x)= sin \;^2 x

- Integrand

f(t)=e^{t^2}

Differentiate:

\;{d}/{d x} ∫↙{ sin \;^2 x}↖{2 sin \;x} e^{t^2} \dt=e^{(2 sin \;x)^2}(2 \cos x)-e^{ ( sin \;^2 x) ^2}(2 sin \;x \cos x)

Now evaluate at

x=π

:

sin \;π=0, \;\; \cos π=-1

So:
-

e^{4 sin \;^2 π}=e^0=1

-

e^{ sin \;^4 π}=e^0=1

-

sin \;π=0

Therefore:

{F^{'}(x) |}_{x=π}=1 ⋅ 2(-1)-1 ⋅ 2(0)(-1)

=-2

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