Computing Definite Integrals — Question 8

Question 8 - Solution

Begin by expanding the integrand: $$e^{x}(1+e^{x})=e^{x}+e^{2x}.$$

Now integrate term by term: $$\int e^{x}\,dx=e^{x}, \qquad \int e^{2x}\,dx=\frac12 e^{2x}.$$

Thus, $$\int e^{x}(1+e^{x})\,dx = e^{x}+\frac12 e^{2x}.$$

Evaluate from $0$ to $\ln 2$: $$\left[e^{x}+\frac12 e^{2x}\right]_{0}^{\ln 2}.$$

Compute the upper bound: $$e^{\ln 2}+\frac12 e^{2\ln 2} = 2+\frac12\cdot 4 = 4.$$

Compute the lower bound: $$e^{0}+\frac12 e^{0}=1+\frac12=\frac32.$$

Subtract: $$4-\frac32=\frac52.$$

$$\boxed{\frac{5}{2}}$$