allery$\displaystyle\int_0^\pi x \sin x\, dx = ?$
A$0$
B$\pi$
C$\pi/2$
D$2\pi$
Answer & Solution
Correct answer: B. $\pi$
**Integration by parts:** $u = x$, $dv = \sin x\, dx$.
$[-x\cos x]_0^\pi + \int_0^\pi \cos x\, dx = \pi + [\sin x]_0^\pi = \pi + 0 = \pi$.
(Or use $\int_0^\pi x f(\sin x)\, dx = (\pi/2)\int_0^\pi f(\sin x)\, dx$ rule.)
Keep practising Unit 6A: Indefinite Integrals & Fundamental Theorem
Browse all Unit 6A: Indefinite Integrals & Fundamental Theorem questions for AP Calculus AB, or explore more Mathematics topics.
Related questions
The integrating factor of the linear differential equation $\dfrac{dy}{dx} + \dfrac{y}{x} The number of arbitrary constants in the general solution of a differential equation of orUsing the substitution $t = 1 + x^2$, $\displaystyle\int \dfrac{x}{1 + x^2}\,dx$ equals:$\displaystyle\int e^x\bigl(f(x) + f'(x)\bigr)\,dx$ equals:$\displaystyle\int \log x\,dx$ equals:$\displaystyle\int ec^2 x\,dx$ equals:$\displaystyle\int \dfrac{1}{ qrt{1 - x^2}}\,dx$ equals:$\displaystyle\int \dfrac{1}{1+x^2}\,dx$ equals: