int_{-1}^{frac{3}{2}}|x sin (pi x)| d x= | vedclass

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

Question

(D) Let $I = \int_{-1}^{3/2} |x \sin(\pi x)| dx$. Since $x \sin(\pi x) \ge 0$ for $x \in [-1, 0]$ and $x \in [1, 3/2]$, and $x \sin(\pi x) \le 0$ for

Vedclass · Accepted Answer

$\frac{3}{\pi}+\frac{1}{\pi^2}$

Vedclass · Answer

(D) Let $I = \int_{-1}^{3/2} |x \sin(\pi x)| dx$. Since $x \sin(\pi x) \ge 0$ for $x \in [-1, 0]$ and $x \in [1, 3/2]$, and $x \sin(\pi x) \le 0$ for $x \in [0, 1]$, we split the integral: $I = \int_{-1}^{0} -x \sin(\pi x) dx + \int_{0}^{1} x \sin(\pi x) dx + \int_{1}^{3/2} -x \sin(\pi x) dx$. Using the property $\int_{-a}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx$ for even functions, note that $f(x) = x \sin(\pi x)$ is an even function. Thus, $I = 2 \int_{0}^{1} x \sin(\pi x) dx - \int_{1}^{3/2} x \sin(\pi x) dx$. Using integration by parts $\int u dv = uv - \int v du$: $\int x \sin(\pi x) dx = -\frac{x \cos(\pi x)}{\pi} + \frac{\sin(\pi x)}{\pi^2}$. Evaluating the first part: $2 \left[ -\frac{x \cos(\pi x)}{\pi} + \frac{\sin(\pi x)}{\pi^2} \right]_{0}^{1} = 2 \left[ (\frac{1}{\pi} + 0) - (0 + 0) \right] = \frac{2}{\pi}$. Evaluating the second part: $-\left[ -\frac{x \cos(\pi x)}{\pi} + \frac{\sin(\pi x)}{\pi^2} \right]_{1}^{3/2} = -\left[ (0 - \frac{1}{\pi^2}) - (\frac{1}{\pi} + 0) \right] = \frac{1}{\pi^2} + \frac{1}{\pi}$. Adding them together: $I = \frac{2}{\pi} + \frac{1}{\pi} + \frac{1}{\pi^2} = \frac{3}{\pi} + \frac{1}{\pi^2}$.

$\int_{-1}^{\frac{3}{2}}|x \sin (\pi x)| d x=$

Similar Questions

One of the possible functions $f(x)$ which satisfies $\int_{-2}^{2} f(x) dx = 0$ is

$ \int_{-3}^{3} \cot^{-1} x \, dx = $

Let $f: R \rightarrow R$ be a continuous function such that $f(x)+f(x+1)=2$ for all $x \in R$. If $I_{1}=\int_{0}^{8} f(x) d x$ and $I_{2}=\int_{-1}^{3} f(x) d x$,then the value of $I_{1}+2 I_{2}$ is equal to:

$\int_0^{\pi / 2} \frac{1}{1+\tan ^{2020}(x)} d x=$

$\int_{-1}^1 \sin ^7 x \cdot \cos ^6 x \, dx = $ . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module