Let A_{1} be the area of the region bounded b | vedclass

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

Question

(A) The curves $y = \sin x$ and $y = \cos x$ intersect at $x = \frac{\pi}{4}$ in the first quadrant.$A_{1}$ is the area bounded by $y = \cos x$,$y = \

Vedclass · Accepted Answer

$A_{1}: A_{2} = 1: \sqrt{2}$ and $A_{1} + A_{2} = 1$

Vedclass · Answer

(A) The curves $y = \sin x$ and $y = \cos x$ intersect at $x = \frac{\pi}{4}$ in the first quadrant.$A_{1}$ is the area bounded by $y = \cos x$,$y = \sin x$ and the $y$-axis from $x = 0$ to $x = \frac{\pi}{4}$:$A_{1} = \int_{0}^{\pi/4} (\cos x - \sin x) dx = [\sin x + \cos x]_{0}^{\pi/4} = (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) - (0 + 1) = \sqrt{2} - 1$.$A_{2}$ is the area bounded by $y = \sin x$ from $x = 0$ to $x = \frac{\pi}{4}$ and $y = \cos x$ from $x = \frac{\pi}{4}$ to $x = \frac{\pi}{2}$:$A_{2} = \int_{0}^{\pi/4} \sin x dx + \int_{\pi/4}^{\pi/2} \cos x dx = [-\cos x]_{0}^{\pi/4} + [\sin x]_{\pi/4}^{\pi/2}$$A_{2} = (-\frac{1}{\sqrt{2}} - (-1)) + (1 - \frac{1}{\sqrt{2}}) = 1 - \frac{1}{\sqrt{2}} + 1 - \frac{1}{\sqrt{2}} = 2 - \frac{2}{\sqrt{2}} = 2 - \sqrt{2} = \sqrt{2}(\sqrt{2} - 1)$.Now,$A_{1} : A_{2} = (\sqrt{2} - 1) : \sqrt{2}(\sqrt{2} - 1) = 1 : \sqrt{2}$.And $A_{1} + A_{2} = (\sqrt{2} - 1) + (2 - \sqrt{2}) = 1$.

Let $A_{1}$ be the area of the region bounded by the curves $y = \sin x$,$y = \cos x$ and the $y$-axis in the first quadrant. Also,let $A_{2}$ be the area of the region bounded by the curves $y = \sin x$,$y = \cos x$,the $x$-axis and $x = \frac{\pi}{2}$ in the first quadrant. Then ..... .

Similar Questions

The figure shows a portion of the graph $y=2x-4x^3$. The line $y=c$ is such that the areas of the regions marked $I$ and $II$ are equal. If $a$ and $b$ are the $x$-coordinates of $A$ and $B$ respectively,then $a+b$ equals:

The area (in square unit) of the region enclosed by the curves $y=x^2$ and $y=x^3$ is

If the area of the region $\{(x, y) : |x-5| \leq y \leq 4 \sqrt{x}\}$ is $A$,then $3A$ is equal to . . . . . . .

The area bounded by the curve $y = \begin{cases} x^2, & x < 0 \\ x, & x \geq 0 \end{cases}$ and the line $y = 4$ is

The area of the region $\{(x, y) : 0 \le y \le 6 - x, y^2 \ge 4x - 3, x \ge 0\}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module