Statement Compute the following definite integrals: (a) ∫013e2x dx\displaystyle\int_0^1 3e^{2x}\,dx∫013e2xdx (b) ∫1e2x dx\displaystyle\int_1^{e}\frac{2}{x}\,dx∫1ex2dx (c) ∫12(x+1x2) dx\displaystyle\int_1^2\Big(x+\frac{1}{x^2}\Big)\,dx∫12(x+x21)dx Solution (a) [32e2x]01=32(e2−1)≈1.5⋅6.389≈9.584\Big[\tfrac32 e^{2x}\Big]_0^1=\tfrac32(e^2-1)\approx 1{.}5\cdot 6{.}389\approx 9{.}584[23e2x]01=23(e2−1)≈1.5⋅6.389≈9.584. (b) [2lnx]1e=2(1−0)=2\big[2\ln x\big]_1^{e}=2(1-0)=2[2lnx]1e=2(1−0)=2. (c) [x22−1x]12=(2−12)−(12−1)=32+12=2\Big[\tfrac{x^2}{2}-\tfrac1x\Big]_1^2=\Big(2-\tfrac12\Big)-\Big(\tfrac12-1\Big)=\tfrac32+\tfrac12=2[2x2−x1]12=(2−21)−(21−1)=23+21=2. a=32(e2−1)≈9.584; b=2; c=2\boxed{a=\tfrac32(e^2-1)\approx 9{.}584;\ b=2;\ c=2}a=23(e2−1)≈9.584; b=2; c=2