Question

$$ int _ { 0 } ^ { 2 } sqrt { x + y + 1 } d x $$

Explanation:

Step1: Use substitution

Let \(u = x + y+1\), then \(du=dx\). When \(x = 0\), \(u=y + 1\); when \(x = 2\), \(u=y + 3\). The integral becomes \(\int_{y + 1}^{y+3}\sqrt{u}du\).

Step2: Integrate \(\sqrt{u}\)

Recall that \(\int u^{n}du=\frac{u^{n + 1}}{n+1}+C\) (\(n
eq- 1\)). For \(n=\frac{1}{2}\), \(\int\sqrt{u}du=\int u^{\frac{1}{2}}du=\frac{2}{3}u^{\frac{3}{2}}+C\).

Step3: Evaluate the definite integral

Using the fundamental theorem of calculus \(\int_{a}^{b}f(u)du=F(b)-F(a)\), where \(F(u)=\frac{2}{3}u^{\frac{3}{2}}\). So \(\frac{2}{3}u^{\frac{3}{2}}\big|_{y + 1}^{y + 3}=\frac{2}{3}(y + 3)^{\frac{3}{2}}-\frac{2}{3}(y + 1)^{\frac{3}{2}}\).

Answer:

\(\frac{2}{3}(y + 3)^{\frac{3}{2}}-\frac{2}{3}(y + 1)^{\frac{3}{2}}\)