Text Compute: (a) ∫01e2x dx\displaystyle\int_{0}^{1}e^{2x}\,dx∫01e2xdx; (b) ∫01dx1+x\displaystyle\int_{0}^{1}\dfrac{dx}{1+x}∫011+xdx. Solution (a) ∫01e2x dx=[e2x2]01=e2−12≈3.1945\displaystyle\int_{0}^{1}e^{2x}\,dx=\left[\dfrac{e^{2x}}{2}\right]_{0}^{1}=\dfrac{e^{2}-1}{2}\approx3.1945∫01e2xdx=[2e2x]01=2e2−1≈3.1945. (b) ∫01dx1+x=[ln(1+x)]01=ln2≈0.6931\displaystyle\int_{0}^{1}\dfrac{dx}{1+x}=\big[\ln(1+x)\big]_{0}^{1}=\ln 2\approx0.6931∫011+xdx=[ln(1+x)]01=ln2≈0.6931. a=e2−12≈3.1945,b=ln2≈0.6931\boxed{a=\dfrac{e^{2}-1}{2}\approx3.1945,\qquad b=\ln 2\approx0.6931}a=2e2−1≈3.1945,b=ln2≈0.6931