Home › JEE Advanced › Mathematics › Definite Integrals › If $\displaystyle f(x) = \int_0^x t \sin t\, dt$…
If $\displaystyle f(x) = \int_0^x t \sin t\, dt$, then $f''(\pi) = ?$
A$-\pi$
B$\pi$
C$0$
D$1$
Answer & Solution
Correct answer: A. $-\pi$
By FTC, $f'(x) = x\sin x$. Differentiating: $f''(x) = \sin x + x\cos x$. At $x = \pi$: $f''(\pi) = \sin\pi + \pi\cos\pi = 0 + \pi(-1) = -\pi$.
Related questions
$\displaystyle\int_0^{\infty} \dfrac{\ln x}{1 + x^2}\, dx = ?$$\displaystyle\int_0^1 \dfrac{x - 1}{\ln x}\, dx = ?$ (Frullani-type)Given $\displaystyle\int_0^{\pi/2} \dfrac{ in^n x}{ in^n x + \cos^n x}\, dx$, the value is$\displaystyle\int_0^{\pi/2} \dfrac{dx}{1 + \tan^3 x} = ?$Evaluate $\displaystyle\lim_{n\to\infty} \dfrac{1^k + 2^k + \cdots + n^k}{n^{k+1}}$ for $k$\displaystyle\int_0^\pi \dfrac{x}{1 + in x}\, dx = ?$$\displaystyle\int_0^{\pi/4} \tan^4 x\, dx = ?$$\displaystyle\int_0^1 \dfrac{x^4(1-x)^4}{1 + x^2}\, dx = ?$ (Putnam — pinpoints $22/7 - \