If [x] is the greatest integer leq x,then pi^ | vedclass

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

Question

(B) Let $I = \pi^{2} \int_{0}^{2} \sin \frac{\pi x}{2} (x-[x])^{[x]} dx$. Since $[x] = 0$ for $x \in [0, 1)$ and $[x] = 1$ for $x \in [1, 2)$,we split

Vedclass · Accepted Answer

$4(\pi-1)$

Vedclass · Answer

(B) Let $I = \pi^{2} \int_{0}^{2} \sin \frac{\pi x}{2} (x-[x])^{[x]} dx$. Since $[x] = 0$ for $x \in [0, 1)$ and $[x] = 1$ for $x \in [1, 2)$,we split the integral: $I = \pi^{2} \left[ \int_{0}^{1} \sin \frac{\pi x}{2} (x-0)^0 dx + \int_{1}^{2} \sin \frac{\pi x}{2} (x-1)^1 dx \right]$ $I = \pi^{2} \left[ \int_{0}^{1} \sin \frac{\pi x}{2} dx + \int_{1}^{2} (x-1) \sin \frac{\pi x}{2} dx \right]$ For the first part: $\int_{0}^{1} \sin \frac{\pi x}{2} dx = [-\frac{2}{\pi} \cos \frac{\pi x}{2}]_0^1 = 0 - (-\frac{2}{\pi}) = \frac{2}{\pi}$. For the second part,use integration by parts: $\int (x-1) \sin \frac{\pi x}{2} dx = (x-1)(-\frac{2}{\pi} \cos \frac{\pi x}{2}) - \int 1 \cdot (-\frac{2}{\pi} \cos \frac{\pi x}{2}) dx = -\frac{2(x-1)}{\pi} \cos \frac{\pi x}{2} + \frac{4}{\pi^2} \sin \frac{\pi x}{2}$. Evaluating from $1$ to $2$: $[-\frac{2(2-1)}{\pi} \cos \pi + \frac{4}{\pi^2} \sin \pi] - [-\frac{2(1-1)}{\pi} \cos \frac{\pi}{2} + \frac{4}{\pi^2} \sin \frac{\pi}{2}] = [\frac{2}{\pi} + 0] - [0 + \frac{4}{\pi^2}] = \frac{2}{\pi} - \frac{4}{\pi^2}$. Summing the parts: $I = \pi^2 [\frac{2}{\pi} + \frac{2}{\pi} - \frac{4}{\pi^2}] = \pi^2 [\frac{4}{\pi} - \frac{4}{\pi^2}] = 4\pi - 4 = 4(\pi-1)$.

If $[x]$ is the greatest integer $\leq x$,then $\pi^{2} \int_{0}^{2}\left(\sin \frac{\pi x}{2}\right)(x-[x])^{[x]} d x$ is equal to :

Similar Questions

If $I_1 = \int_0^{3 \pi} f(\cos^2 x) dx$ and $I_2 = \int_0^\pi f(\cos^2 x) dx$, then

$\int_0^{\pi / 2} \frac{d x}{1+\tan ^3 x}$ is equal to :

Evaluate the definite integral $\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} x}{\cos ^{2} x+4 \sin ^{2} x} d x$.

The value of the integral $I = \int_{\frac{\pi}{24}}^{\frac{5\pi}{24}} \frac{dx}{1+\sqrt[3]{\tan 2x}}$ is:

The value of the integral $\sum\limits_{k = 1}^n {\int_0^1 {f(k - 1 + x)\,dx} } $ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module