Solve definite integral by substitution

Question

I have to solve $\int_{-1}^{1} \dfrac{x}{\sqrt{5-4x}} \,dx $ by using substitution. So let $t = 5 - 4x$. $\varphi(t) = \dfrac{5-t}{4}$. Consequently $\varphi(t) = x$. So $-1 \leq \varphi(t) \leq 1$ and $1 \leq t \leq 9$. So we can now substitute $x$ with $\varphi(t)$.

$d\varphi'(t) = x' = 1$. $dt = -4dx$ consequently $dx = -dt/4$. So now we should solve $-\int_{1}^{9} {\dfrac{\varphi(t)}{\sqrt{5-4\varphi(t)}}\varphi'(t)\dfrac{dt}{4}} = -\int_{1}^{9} {\dfrac{5-t}{4\sqrt{t}}\dfrac{dt}{4}} = -\dfrac{1}{16}\left[ 5\int_{1}^{9} {\sqrt{t}^{-1} dt} - \int_{1}^{9} {\sqrt{t} dt}\right] = -\dfrac{1}{16}\left[5*2\sqrt{t}\bigg|_{1}^{9} - \dfrac{2\sqrt{t}^3}{3}\bigg|_{1}^{9}\right] = -\dfrac{1}{16} \left[ 10*3 - 10*1 - 2*27/3 + 2*1/3 \right] = -\dfrac{30-10-18+2/3}{16} = -1/6$.

But the result should be 1/6 (thus not negative). Where is my error? Am I misusing the theorem of variable substitution? Thanks in advance.

Answer 1

Note that $\varphi$ is decreasing. So, after the substitution, your integral should be $\int_9^1$, instead of $\int_1^9$.

Answer 2

It's cleaner to eliminate the radicals

Let $t=\sqrt{5-4x}\implies dt=-\dfrac{4dx}{2\sqrt{5-4x}}$

$$I=-\dfrac12\int_9^1(5-t^2)dt=?$$