Text Evaluate the following immediate definite integrals: a) ∫02(2x3−x) dx\int_0^2(2x^3-x)\,dx∫02(2x3−x)dx b) ∫0ln2e2x dx\int_0^{\ln 2}e^{2x}\,dx∫0ln2e2xdx c) ∫0π/3cosx dx\int_0^{\pi/3}\cos x\,dx∫0π/3cosxdx d) ∫12dxx2\int_1^2\dfrac{dx}{x^2}∫12x2dx e) ∫01e−x dx\int_0^1 e^{-x}\,dx∫01e−xdx f) ∫−11(4−x2) dx\int_{-1}^1(4-x^2)\,dx∫−11(4−x2)dx Solution a) [x42−x22]02=8−2=6.\left[\dfrac{x^4}{2}-\dfrac{x^2}{2}\right]_0^2=8-2=6.[2x4−2x2]02=8−2=6. b) [e2x2]0ln2=4−12=32.\left[\dfrac{e^{2x}}{2}\right]_0^{\ln 2}=\dfrac{4-1}{2}=\dfrac32.[2e2x]0ln2=24−1=23. c) [sinx]0π/3=32≈0.8660.\big[\sin x\big]_0^{\pi/3}=\dfrac{\sqrt3}{2}\approx0.8660.[sinx]0π/3=23≈0.8660. d) [−1x]12=−12+1=12.\left[-\dfrac1x\right]_1^2=-\dfrac12+1=\dfrac12.[−x1]12=−21+1=21. e) [−e−x]01=1−1e≈0.6321.\big[-e^{-x}\big]_0^1=1-\dfrac1e\approx0.6321.[−e−x]01=1−e1≈0.6321. f) [4x−x33]−11=(4−13)−(−4+13)=223≈7.333.\left[4x-\dfrac{x^3}{3}\right]_{-1}^1=\left(4-\dfrac13\right)-\left(-4+\dfrac13\right)=\dfrac{22}{3}\approx7.333.[4x−3x3]−11=(4−31)−(−4+31)=322≈7.333. a=6, b=32, c=32, d=12, e=1−1e, f=223\boxed{a=6,\ b=\tfrac32,\ c=\tfrac{\sqrt3}{2},\ d=\tfrac12,\ e=1-\tfrac1e,\ f=\tfrac{22}{3}}a=6, b=23, c=23, d=21, e=1−e1, f=322