# Thread: [SOLVED] U Substitution Integral

1. ## [SOLVED] U Substitution Integral

What is the $\int_1^9 \frac{1}{\sqrt{x}} e^{-\sqrt{x}} dx$ ?

The answer should be $2e^{-1} - 2e^{-3}$ .

How do I use u substitution?

Thanks for helping!!

2. Originally Posted by juicysharpie
What is the $\int_1^9 of \frac{1}{\sqrt{x}} e^{-\sqrt{x}} dx$ ?

The answer should be $2e^{-1} - 2e^{-3}$

How do I use u substitution?

Thanks for helping!!
Let $u = - \sqrt{x}$

this would mean $du = - \frac 1{\sqrt{x}}~dx$, or in other words, $-du = \frac 1{\sqrt x}~dx$

can you finish?

3. Originally Posted by juicysharpie
What is the integral from 1 to 9 of $(\frac{1}{\sqrt{x}}) (e^-\sqrt{x}) (dx)$ ?

The answer should be $2e^{-1} - 2e^{-3}$

How do I use u substitution?

Thanks for helping!!
$u = -\sqrt{x}$

$du = -\frac{1}{2\sqrt{x}} \, dx$

$\int_1^9 \frac{1}{\sqrt{x}} \cdot e^{-\sqrt{x}} \, dx$

$-2 \int_1^9 -\frac{1}{2\sqrt{x}} \cdot e^{-\sqrt{x}} \, dx$

$-2 \int_{-1}^{-3} e^u \, du$

$2 \int_{-3}^{-1} e^u \, du$

$2(e^{-1} - e^{-3})$