∫-1 to 1 f(x)dx, where f(x) = {x + 1, x ≤ 0; 2x, x > 0}

Explanation:

Step1: Split the integral

We split $\int_{-1}^{1}f(x)dx$ into $\int_{-1}^{0}f(x)dx+\int_{0}^{1}f(x)dx$. Since $f(x)=x + 1$ for $x\leq0$ and $f(x)=2x$ for $x>0$. So we have $\int_{-1}^{0}(x + 1)dx+\int_{0}^{1}2xdx$.

Step2: Integrate $\int_{-1}^{0}(x + 1)dx$

Using the power - rule of integration $\int x^n dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$, we get $\int_{-1}^{0}(x + 1)dx=\left[\frac{x^{2}}{2}+x\right]_{-1}^{0}$. [ \begin{align*} \left(\frac{0^{2}}{2}+0\right)-\left(\frac{(-1)^{2}}{2}-1\right)&=0-\left(\frac{1}{2}-1\right)\ &=0-\frac{1}{2}+1\ &=\frac{1}{2} \end{align*} ]

Step3: Integrate $\int_{0}^{1}2xdx$

Using the power - rule $\int 2x dx=x^{2}+C$, we get $\int_{0}^{1}2xdx=\left[x^{2}\right]_{0}^{1}=1^{2}-0^{2}=1$.

Step4: Sum the two results

$\int_{-1}^{1}f(x)dx=\frac{1}{2}+1=\frac{3}{2}$.

Answer:

$\frac{3}{2}$