Evaluating an integral by using du/dx
Matthew Harrington 
I am having trouble evaluating this formula. I have tried to solve it and got 1/4 as an answer. However by using the function Sf(x)dx in the Calc button in the TI-84 caclulator, I get 3.7037037 Which means my answer is wrong.
Here is the formuala that needs to be evaluated:
$\int_1^5 \frac {x}{\sqrt{3x+1}} dx $
one a side note I have : $$u = 3x+1$$ $$du/dx = 3$$ $$du = 3dx$$ $$x=(1/3)(u-1)$$ $$ \begin{array}{c|lcr} x & u \\ \hline 5 & 16 \\ 1 & 4 \\ \end{array} $$
$\int_1^5 (x)(3x+1)^{-.5} dx $
(1/3)$\int_{16}^4 (1/3)(u-1)(u)^{-.5} du$
$\endgroup$1 Answer
$\begingroup$There are many ways to do this. One way is as follows: Let $\sqrt{3x+1} = t$, i.e., $3x+1 = t^2$, i.e., $x = \dfrac{t^2-1}3$. We then have $dx = \dfrac{2t dt}3$. Also, note that when $x=1$, we get $t=2$ and when $x=5$, we get $t=4$.
Hence, \begin{align} \int_1^5 \dfrac{x}{\sqrt{3x+1}}dx & = \int_2^{4} \dfrac{t^2-1}{3t} \dfrac{2t dt}3 = \dfrac29 \int_2^4 (t^2-1) dt = \dfrac29 \left(\dfrac{t^3}3 - t\right)_{t=2}^{t=4}\\ & = \dfrac29 \left(\dfrac{64}3 - 4 - \dfrac83 + 2 \right) = \dfrac29 \left(\dfrac{56}3-2\right) = \color{red}{\dfrac{100}{27}} \end{align}
EDIT
As per your substitution, i.e., $3x+1 = u$, we get $x = \dfrac{u-1}3$, i.e., $dx = \dfrac{du}3$. Also, as you have noted when $x=1$, we have $u = 4$ and $x=5$, gives us $u=16$. Hence, the integral becomes \begin{align} \int_1^5 \dfrac{x}{\sqrt{3x+1}}dx & = \int_4^{16} \dfrac{u-1}{3 \sqrt{u}} \dfrac{du}3 = \dfrac19 \int_4^{16} \left(\sqrt{u} - \dfrac1{\sqrt{u}}\right)du = \dfrac19 \left(\dfrac{u^{3/2}}{3/2}-\dfrac{u^{1/2}}{1/2}\right)_{u=4}^{u=16}\\ & = \dfrac29\left(\dfrac{16^{3/2}}3 - 16^{1/2} - \dfrac{4^{3/2}}3 + 4^{1/2}\right) = \dfrac29\left(\dfrac{64-8}3-2\right) = \color{red}{\dfrac{100}{27}} \end{align}
$\endgroup$ 2
