It depends on what your substitution is. for instance, if you used the substitution $\displaystyle u = \sqrt{x^3 + 3}$, then you'd just plug in
when $\displaystyle x = 5$, then $\displaystyle u = \sqrt{5^3 + 3} = \sqrt{128} = u(5)$
when $\displaystyle x = -1$, then $\displaystyle u = \sqrt{(-1)^3 + 3} = \sqrt{2} = u(-1)$
i have no idea.
the integral you are using is $\displaystyle \frac {10}3 \int 1~du$, I assume
then you should get $\displaystyle \frac {10u}3 \bigg|_{\sqrt{2}}^{\sqrt{128}} = \frac {10 \sqrt{128}}3 - \frac {10 \sqrt{2}}3 = 32.9983$
maybe you should start from the beginning. i think you're mixing yourself up in the middle somewhere. what substitution are you using?
Hello,
Ok, now calculate $\displaystyle \frac{du}{dx}$
The derivative for such a thing is $\displaystyle \frac{u'(x)}{2 \sqrt{u(x)}}$
Thus $\displaystyle \frac{du}{dx}=\frac{3x^2}{2\sqrt{x^3+3}} \Longrightarrow dx=du \frac{2\sqrt{x^3+3}}{3x^2}=du \frac{2 u}{3x^2}$
Now replacing in the integral :
$\displaystyle \int_{-1} ^5\;\frac{5x^2}{\sqrt{x^3+3}}\;dx=\int_{-1}^5 \frac{5x^2}{u} dx$
Replacing dx :
$\displaystyle \int_{-1}^5 \frac{5{\color{red}x^2}}{\color{blue}u} \frac{2 {\color{blue}u}}{3{\color{red}x^2}} du= \dots$
Okay. so let's get the limits out of the way.
we already saw that the new limits for u are $\displaystyle \sqrt{128}$ and $\displaystyle \sqrt{2}$. so good.
Now, $\displaystyle u = \sqrt{x^3 + 3}$
$\displaystyle \Rightarrow du = \frac {3x^2}{2 \sqrt{x^3 + 3}}~dx$ .............(by the chain rule)
$\displaystyle \Rightarrow \frac 23 ~du = \frac {x^2}{\sqrt{x^3 + 3}}~dx$
$\displaystyle \Rightarrow \frac {10}3~du = \frac {5x^2}{\sqrt{x^3 + 3}}~dx$ ..........................multiplied both sides by 5
What Luck! the right hand side is exactly our integral. this does not always happen, but it is really nice here, and we won't complain. so we can simply replace our entire integral with 10/3 in du.
So, our integral becomes:
$\displaystyle \int_{\sqrt{2}}^{\sqrt{128}} \frac {10}3~du = \frac {10}3u \bigg|_{\sqrt{2}}^{\sqrt{128}} = \frac {10}3 (\sqrt{128} - \sqrt{2})$
and then just plug that into your calculator
do you get it?