int_{,pi }^{,10pi } {|sin x|dx} is | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) We know that the function $f(x) = |\sin x|$ is periodic with period $\pi$. Using the property $\int_{a}^{a+nT} f(x) dx = n \int_{0}^{T} f(x) dx$,w

Vedclass · Accepted Answer

$18$

Vedclass · Answer

(D) We know that the function $f(x) = |\sin x|$ is periodic with period $\pi$. Using the property $\int_{a}^{a+nT} f(x) dx = n \int_{0}^{T} f(x) dx$,where $T$ is the period,we have: $\int_{\,\pi }^{\,10\pi } {|\sin x|dx} = \int_{\,0}^{\,9\pi } {|\sin x|dx}$ (by shifting the limits by $-\pi$). Since the period is $\pi$,we can write this as: $9 \int_{\,0}^{\,\pi } {|\sin x|dx} = 9 \int_{\,0}^{\,\pi } {\sin x dx}$ (since $\sin x \ge 0$ in $[0, \pi]$). Evaluating the integral: $9 [-\cos x]_{0}^{\pi} = 9 (-(\cos \pi) + \cos 0) = 9 (-(-1) + 1) = 9(1 + 1) = 18$.

$\int_{\,\pi }^{\,10\pi } {|\sin x|dx}$ is

Similar Questions

$\int_{0}^{1} \log \left(\frac{1}{x}-1\right) d x$ is equal to

If $f(x) = f(\Pi + e - x)$ and $\int_{e}^{\Pi} f(x) dx = \frac{2}{e + \Pi}$,then $\int_{e}^{\Pi} x f(x) dx$ is equal to

The value of $\sum_{r=1}^{20} \left( \sqrt{\pi \left( \int_0^r x |\sin \pi x| dx \right)} \right)$ is . . . . . . .

$\int_{1}^{6\pi}([\sec^{-1}x]+[\cot^{-1}x])dx$ is equal to (where $[.]$ denotes the greatest integer function).

If $\int_{ - a}^a {\sqrt {\frac{{a - x}}{{a + x}}} \,dx = k\pi ,} $ then $k = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module