Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
If $x\in[0,5]$ then $u\in[1,6]$. I'm not sure whether that's a typo or actually your mistake. –  Matt Pressland Dec 11 '12 at 17:47
    
...$0$ transforms to $1$. Alternatively transform back into $x$ and use the original $[0,5]$. –  Jp McCarthy Dec 11 '12 at 17:48
    
Thats a typo, sorry. It is supposed to be [1,6] –  Lizi Dec 11 '12 at 17:48
    
What are you expecting to get? 6.2223, perhaps? –  Rick Decker Dec 11 '12 at 17:53
1  
I figured it out. It was just a calculation error. –  Lizi Dec 11 '12 at 17:54
add comment

2 Answers

up vote 0 down vote accepted

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 $7.9981$.

share|improve this answer
add comment

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}

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.