Question

Evaluate: $\int\sqrt{7x-10-x^{2}}dx$

• A. $\frac{1}{4}\left(2x-7\right)\sqrt{7x-10-x^{2}}+\frac{9}{8}sin^{-1}\left(\frac{2x-7}{3 }\right)+C $
• B. $-\frac{1}{4}\left(2x-7\right)\sqrt{7x-10-x^{2}}-\frac{9}{8}sin^{-1}\left(\frac{2x-7}{3 }\right)+C $
• C. $-\frac{1}{4}\left(2x-7\right)\sqrt{7x-10-x^{2}}+\frac{9}{8}sin^{-1}\left(\frac{2x-7}{3 }\right)+C $
• D. None of these

Answer 1

Answer: (A)

$I= \int\sqrt{-\left(x^{2}-7x +10\right)}dx $

$I \int\sqrt{-\left(x^{2}-7x+\frac{49}{4}-\frac{49}{4}+10\right)}dx $

$\Rightarrow I = \int\sqrt{-\left\{\left(x-\frac{7}{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}\right\}}dx = \int\sqrt{\left(\frac{3}{2}\right)^{2}-\left(x-\frac{7}{2}\right)^{2}}dx$

$\Rightarrow I= \frac{1}{2}\left(x-\frac{7}{2}\right)\sqrt{\left(\frac{3}{2}\right)^{2}-\left(x-\frac{7}{2}\right)^{2}} +\frac{1}{2}\cdot\left(\frac{3}{2}\right)^{2} sin^{-1}\left(\frac{x-\left(\frac{7}{2}\right)}{\frac{3}{2}}\right)+C $

$\Rightarrow I =\frac{1}{4}\left(2x-7\right)\sqrt{7x-10-x^{2}}+\frac{9}{8}sin^{-1}\left(\frac{2x-7}{3 }\right)+C $