int_0^{10} (5 - sqrt{10x - x^2}) , dx = | vedclass

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

Question

(C) Let $I = \int_0^{10} (5 - \sqrt{10x - x^2}) \, dx$. First, complete the square inside the square root: $10x - x^2 = -(x^2 - 10x) = -(x^2 - 10x + 2

Vedclass · Accepted Answer

$\frac{1}{2}(100 - 25\pi)$

Vedclass · Answer

(C) Let $I = \int_0^{10} (5 - \sqrt{10x - x^2}) \, dx$. First, complete the square inside the square root: $10x - x^2 = -(x^2 - 10x) = -(x^2 - 10x + 25 - 25) = 25 - (x - 5)^2$. So, $I = \int_0^{10} 5 \, dx - \int_0^{10} \sqrt{5^2 - (x - 5)^2} \, dx$. The first part is $\int_0^{10} 5 \, dx = [5x]_0^{10} = 50$. For the second part, use the formula $\int \sqrt{a^2 - u^2} \, du = \frac{u}{2}\sqrt{a^2 - u^2} + \frac{a^2}{2}\sin^{-1}(\frac{u}{a})$. Here $u = x - 5$, so $du = dx$. When $x=0, u=-5$; when $x=10, u=5$. $\int_{-5}^5 \sqrt{5^2 - u^2} \, du = [\frac{u}{2}\sqrt{25 - u^2} + \frac{25}{2}\sin^{-1}(\frac{u}{5})]_{-5}^5$. $= (0 + \frac{25}{2}\sin^{-1}(1)) - (0 + \frac{25}{2}\sin^{-1}(-1)) = \frac{25}{2}(\frac{\pi}{2}) - \frac{25}{2}(-\frac{\pi}{2}) = \frac{25\pi}{4} + \frac{25\pi}{4} = \frac{25\pi}{2}$. Thus, $I = 50 - \frac{25\pi}{2} = \frac{100 - 25\pi}{2} = \frac{1}{2}(100 - 25\pi)$.

$\int_0^{10} (5 - \sqrt{10x - x^2}) \, dx = $

Similar Questions

$\int_0^{\pi /2} {\frac{{\sin x\cos x\,dx}}{{{{\cos }^2}x + 3\cos x + 2}}} = $

Evaluate the definite integral $\int_{0}^{\frac{\pi}{2}} \cos 2x \, dx$.

If $\int_a^b {x^3} dx = 0$ and $\int_a^b {x^2} dx = \frac{2}{3}$,then the values of $a$ and $b$ are respectively:

$\int_{0}^{\pi} \sqrt{\frac{1 + \cos 2x}{2}} \, dx$ is equal to

If $[x]$ denotes the greatest integer function of $x$ and $\int_{-\frac{3}{2}}^{\frac{3}{2}}[2x-3] dx = k$, then $\left|k+\frac{1}{2}\right| = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module