Question

evaluate the integral
int_{-3}^{0} \frac{dx}{sqrt{1 - x}}

Explanation:

Step1: Use substitution

Let $u = 1 - x$, then $du=-dx$. When $x=-3$, $u = 1-(-3)=4$; when $x = 0$, $u=1 - 0 = 1$.

Step2: Rewrite the integral

The integral $\int_{-3}^{0}\frac{dx}{\sqrt{1 - x}}$ becomes $-\int_{4}^{1}\frac{du}{\sqrt{u}}=\int_{1}^{4}u^{-\frac{1}{2}}du$.

Step3: Apply the power - rule for integration

The power - rule for integration is $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n
eq - 1$). For $n=-\frac{1}{2}$, we have $\int u^{-\frac{1}{2}}du=\frac{u^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+C = 2u^{\frac{1}{2}}+C$.

Step4: Evaluate the definite integral

$2u^{\frac{1}{2}}\big|_{1}^{4}=2\sqrt{4}-2\sqrt{1}$.
$2\times2 - 2\times1=4 - 2$.

Answer:

$2$