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Okay, so I'm working on definite integrals and I calculated the indefinte integral of $\frac{x}{\sqrt{x+1}}$ to be $\frac{2}{3}u^{3/2}-2\sqrt{u}$ where $u=x+1$. The definite integral is on the interval $[0,5]$ so I used my $u$-subsitution equation on the interval $[1,6]$. I keep getting $7.9981$ but I know that's incorrect. What am I doing wrong?

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    I figured it out. It was just a calculation error.2012-12-11

2 Answers 2

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Using the substitution $u=x+1$, you get $\int\frac{xdx}{\sqrt{x+1}}=\int\frac{u-1}{\sqrt{u}}du=\int(u^\frac{1}{2}-u^{-\frac{1}{2}})du=\frac{2}{3}u^\frac{3}{2}-2\sqrt{u}$as you did. Then, to calculate the definite integral, you plug in the following: $\int_1^5\frac{u-1}{\sqrt{u}}du=\left[\frac{2}{3}u^\frac{3}{2}-2\sqrt{u}\right]_1^6=\frac{4}{3}+2\sqrt{6}=6.23231\cdots$not $7.9981$.

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Same answer, but with the useful trick of "adding $0$":

\begin{align} \int \frac{x}{\sqrt{x+1}}dx = \int \frac{x+1 -1}{\sqrt{x+1}}&= \int \left(\frac{x+1}{\sqrt{x+1}} - \frac{1}{\sqrt{x+1}} \right)dx \\ &= \int\left( (x+1)^{\frac{1}{2}} - (x+1)^{- \frac{1}{2}}\right)dx \\ &= \frac{2}{3} (x+1)^{\frac{3}{2}} - 2 (x+1)^{\frac{1}{2}}. \end{align}