int_{1/4}^{1/2} frac{dx}{sqrt{x - x^2}} = | vedclass

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

Question

(D) Let $I = \int_{1/4}^{1/2} \frac{dx}{\sqrt{x - x^2}}$.We can rewrite the expression inside the square root by completing the square:$x - x^2 = -(x^

Vedclass · Accepted Answer

$\frac{\pi}{6}$

Vedclass · Answer

(D) Let $I = \int_{1/4}^{1/2} \frac{dx}{\sqrt{x - x^2}}$.We can rewrite the expression inside the square root by completing the square:$x - x^2 = -(x^2 - x) = -((x - 1/2)^2 - 1/4) = (1/2)^2 - (x - 1/2)^2$.Thus,$I = \int_{1/4}^{1/2} \frac{dx}{\sqrt{(1/2)^2 - (x - 1/2)^2}}$.Using the standard integral formula $\int \frac{dx}{\sqrt{a^2 - u^2}} = \sin^{-1}(\frac{u}{a}) + C$,we get:$I = [\sin^{-1}(\frac{x - 1/2}{1/2})]_{1/4}^{1/2} = [\sin^{-1}(2x - 1)]_{1/4}^{1/2}$.Evaluating at the limits:$I = \sin^{-1}(2(1/2) - 1) - \sin^{-1}(2(1/4) - 1) = \sin^{-1}(0) - \sin^{-1}(-1/2)$.$I = 0 - (-\pi/6) = \pi/6$.

$\int_{1/4}^{1/2} \frac{dx}{\sqrt{x - x^2}} = $

Similar Questions

$\int_2^3 \frac{d x}{x^2-x}$ is equal to

The value of $b > 3$ for which $12 \int \limits_{3}^{b} \frac{1}{(x^{2}-1)(x^{2}-4)} dx = \log _{e}(\frac{49}{40})$ is equal to

$\int_{-\pi}^{\frac{\pi}{2}} \sin x \cdot \sin^2(\cos x) \, dx =$

Let $l = \mathop {Lim}\limits_{x \to \infty } \int\limits_x^{2x} \frac{dt}{t}$ and $m = \mathop {Lim}\limits_{x \to \infty } \frac{1}{x \ln x} \int\limits_1^x \ln t \, dt$. Then the correct statement is:

The value of $\int_{-1}^{2} \frac{|x|}{x} d x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module