int_{-4}^5 frac{1}{sqrt{20+x-x^2}} dx= | vedclass

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

Question

(C) To solve the integral $I = \int_{-4}^5 \frac{1}{\sqrt{20+x-x^2}} dx$, first complete the square for the quadratic expression inside the square roo

Vedclass · Accepted Answer

$\pi$

Vedclass · Answer

(C) To solve the integral $I = \int_{-4}^5 \frac{1}{\sqrt{20+x-x^2}} dx$, first complete the square for the quadratic expression inside the square root: $20+x-x^2 = -(x^2-x-20) = -(x^2-x+\frac{1}{4}-\frac{1}{4}-20) = -((x-\frac{1}{2})^2 - \frac{81}{4}) = \frac{81}{4} - (x-\frac{1}{2})^2$. Now the integral becomes $I = \int_{-4}^5 \frac{1}{\sqrt{(\frac{9}{2})^2 - (x-\frac{1}{2})^2}} dx$. Using the standard integral formula $\int \frac{1}{\sqrt{a^2-u^2}} du = \sin^{-1}(\frac{u}{a}) + C$, we get: $I = [\sin^{-1}(\frac{x-1/2}{9/2})]_{-4}^5 = [\sin^{-1}(\frac{2x-1}{9})]_{-4}^5$. Evaluating at the limits: $I = \sin^{-1}(\frac{2(5)-1}{9}) - \sin^{-1}(\frac{2(-4)-1}{9}) = \sin^{-1}(\frac{9}{9}) - \sin^{-1}(\frac{-9}{9}) = \sin^{-1}(1) - \sin^{-1}(-1)$. Since $\sin^{-1}(1) = \frac{\pi}{2}$ and $\sin^{-1}(-1) = -\frac{\pi}{2}$, we have: $I = \frac{\pi}{2} - (-\frac{\pi}{2}) = \pi$.

$\int_{-4}^5 \frac{1}{\sqrt{20+x-x^2}} dx=$

Similar Questions

$\int_0^a \frac{x-a}{x+a} dx =$

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

$\int\limits_0^\infty [2e^{-x}]\, dx$,where $[x]$ denotes the greatest integer function,is equal to:

$\int_0^{\pi / 2} \frac{\pi \sin x}{1+\cos ^2 x} d x$ is equal to

The value of the integral $\int_0^2 \frac{\sqrt{x}(x^2 + x + 1)}{(\sqrt{x}+1)(\sqrt{x^4+x^2+1})} dx$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module