Let $f:[-2, 3] \to [0, \infty)$ be a continuous function such that $f(1-x) = f(x)$ for all $x \in [-2, 3]$. If $R_1$ is the numerical value of the area of the region bounded by $y = f(x)$,$x = -2$,$x = 3$ and the $x$-axis,and $R_2 = \int_{-2}^3 x f(x) dx$,then:
- A$3R_1 = 2R_2$
- B$2R_1 = 3R_2$
- C$R_1 = R_2$
- D$R_1 = 2R_2$
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If $I_n = \int_{-\pi}^{\pi} \frac{\sin(nx)}{(1+\pi^x) \sin x} dx$,$n=0, 1, 2, \ldots$,then
$(A)$ $I_n = I_{n+2}$
$(B)$ $\sum_{m=1}^{10} I_{2m+1} = 10\pi$
$(C)$ $\sum_{m=1}^{10} I_{2m} = 0$
$(D)$ $I_n = I_{n+1}$
$(A)$ $I_n = I_{n+2}$
$(B)$ $\sum_{m=1}^{10} I_{2m+1} = 10\pi$
$(C)$ $\sum_{m=1}^{10} I_{2m} = 0$
$(D)$ $I_n = I_{n+1}$
DifficultIIT 2009
View SolutionThe value of $\sum_{n=1}^{10} \int_{-2n-1}^{-2n} \sin^{27} x \, dx + \sum_{n=1}^{10} \int_{2n}^{2n+1} \sin^{27} x \, dx$ is equal to
MediumWBJEE 2020
View Solution$\int_{-\pi}^{\pi} x^2 \sin x \, dx =$
$\int_0^{\frac{\pi}{2}} \left( \frac{\sqrt[n]{\sec x}}{\sqrt[n]{\sec x} + \sqrt[n]{\operatorname{cosec} x}} \right) dx = $
$\int_0^{\pi /2} \frac{x \sin x \cos x}{\cos^4 x + \sin^4 x} \, dx = $
DifficultIIT 1985
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