Area bounded by the curve y = min {sin^2x, co | vedclass

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

Question

(C) The curve $y = \min \{\sin^2x, \cos^2x \}$ changes its definition at points where $\sin^2x = \cos^2x$,i.e.,$	an^2x = 1$,which gives $x = \frac{\p

Vedclass · Accepted Answer

$\frac{5(\pi - 2)}{8}$ square units

Vedclass · Answer

(C) The curve $y = \min \{\sin^2x, \cos^2x \}$ changes its definition at points where $\sin^2x = \cos^2x$,i.e.,$	an^2x = 1$,which gives $x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}$.In the interval $[0, \frac{\pi}{4}]$,$\sin^2x \leq \cos^2x$,so $y = \sin^2x$.In the interval $[\frac{\pi}{4}, \frac{3\pi}{4}]$,$\cos^2x \leq \sin^2x$,so $y = \cos^2x$.In the interval $[\frac{3\pi}{4}, \frac{5\pi}{4}]$,$\sin^2x \leq \cos^2x$,so $y = \sin^2x$.The total area $A$ is given by:$A = \int_{0}^{\frac{\pi}{4}} \sin^2x \, dx + \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \cos^2x \, dx + \int_{\frac{3\pi}{4}}^{\frac{5\pi}{4}} \sin^2x \, dx$.Using $\sin^2x = \frac{1-\cos 2x}{2}$ and $\cos^2x = \frac{1+\cos 2x}{2}$:$A = \int_{0}^{\frac{\pi}{4}} \frac{1-\cos 2x}{2} dx + \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{1+\cos 2x}{2} dx + \int_{\frac{3\pi}{4}}^{\frac{5\pi}{4}} \frac{1-\cos 2x}{2} dx$.$A = \left[ \frac{x}{2} - \frac{\sin 2x}{4} ight]_{0}^{\frac{\pi}{4}} + \left[ \frac{x}{2} + \frac{\sin 2x}{4} ight]_{\frac{\pi}{4}}^{\frac{3\pi}{4}} + \left[ \frac{x}{2} - \frac{\sin 2x}{4} ight]_{\frac{3\pi}{4}}^{\frac{5\pi}{4}}$.$A = (\frac{\pi}{8} - \frac{1}{4}) + ((\frac{3\pi}{8} - \frac{1}{4}) - (\frac{\pi}{8} + \frac{1}{4})) + ((\frac{5\pi}{8} - 0) - (\frac{3\pi}{8} + \frac{1}{4}))$.$A = \frac{\pi}{8} - \frac{1}{4} + \frac{2\pi}{8} - \frac{2}{4} + \frac{2\pi}{8} - \frac{1}{4} = \frac{5\pi}{8} - 1 = \frac{5\pi - 8}{8}$.

Area bounded by the curve $y = \min \{\sin^2x, \cos^2x \}$ and $x-$ axis between the ordinates $x = 0$ and $x = \frac{5\pi}{4}$ is

Similar Questions

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

The curve $y = ax^2 + bx + c$ passes through the point $(1, 2)$ and its tangent at the origin is the line $y = x$. The area bounded by the curve,the ordinate of the curve at its minima,and the tangent line is

The area (in sq units) bounded by the curves $y^2=4x$ and $x^2=4y$ is

If the area of the region bounded by the curves $y = x^2$,$y = \frac{1}{x}$ and the lines $y = 0$ and $x = t$ $(t > 1)$ is $1 \, \text{sq. unit}$,then $t$ is equal to

If $f(x)$ is a continuous,increasing,and odd function such that $\int_{-1}^{4} f(x) \,dx = 10$ and $\int_{0}^{1} f(x) \,dx = \frac{3}{2}$,then the area bounded by $y = f(x)$,the $x$-axis,and the ordinates $x = -4$ and $x = 4$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module