Use Part 2 of the Fundamental Theorem of Calculus to find:
Solution (a) An antiderivative of f(x)=x2 is F(x)=x33. By Part 2 of the Fundamental Theorem of Calculus, ∫1−2x2dx=[x33]1−2=133−(−2)33=13+83=93=3
(b) An antiderivative of f(x)=cosx is F(x)=sinx. By Part 2 of the Fundamental Theorem of Calculus, ∫π/60cosxdx=[sinx]π/60=sinπ6−sin0=12
(c) An antiderivative of f(x)=1√1−x2 is F(x)=sin−1x, provided |x|<1. By Part 2 of the Fundamental Theorem of Calculus, ∫√3/201√1−x2dx=[sin−1x]√3/20=sin−1√32−sin−10=π3−0=π3
(d) An antiderivative of f(x)=1x is F(x)=ln|x|. By Part 2 of the Fundamental Theorem of Calculus, ∫211xdx=[ln|x|]21=ln2−ln1=ln2−0=ln2