The maximum value of $f(x)=\int_0^1 t \sin (x+\pi t) d t$ is

$\frac{1}{2 \pi^2} \sqrt{\pi^2+4}$

JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

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The maximum value of

f(x)=\int_0^1 t \sin (x+\pi t) d t

View answer and explanation

Correct option: B

\frac{1}{\pi^2} \sqrt{\pi^2+4}

Option B,

\frac{1}{\pi^2} \sqrt{\pi^2+4}

, is correct because it is the maximum value obtained by evaluating the integral and then finding the maximum of the resulting trigonometric function.

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