Gujarat BoardEnglish MediumSTD 12 ScienceMathsIntegrals1 Mark
Question
Evaluate $\int_{4}^{9} \frac{\sqrt{x}}{\left(30-x^{\frac{3}{2}}\right)^{2}} d x$
✓
Answer
Let $I=\int_{4}^{9} \frac{\sqrt{x}}{\left(30-x^{\frac{3}{2}}\right)^{2}} d x$. We first find the antiderivative of the integrand. Put $30-x^{\frac{3}{2}}=t$. Then $-\frac{3}{2} \sqrt{x} d x=d t$ or $\sqrt{x} d x=-\frac{2}{3} d t$ Thus, $\int \frac{\sqrt{x}}{\left(30-x^{\frac{3}{2}}\right)^{2}} d x$ = $-\frac{2}{3} \int \frac{d t}{t^{2}}=\frac{2}{3}\left[\frac{1}{t}\right]=\frac{2}{3}\left[\frac{1}{\left(30-x^{\frac{3}{2}}\right)}\right]=\mathrm{F}(x)$ Therefore, by the second fundamental theorem of calculus, we have I = F(9) - F(4) = $\frac{2}{3}\left[\frac{1}{(30-27)}-\frac{1}{30-8}\right]$ $=\frac{2}{3}\left[\frac{1}{3}-\frac{1}{22}\right]=\frac{19}{99}$
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