Using Part 2 of the Fundamental Theorem of Calculus

Use Part 2 of the Fundamental Theorem of Calculus to find:

1. \({\int_{-2}^{1}{x^{2} dx}}\)
2. \({\int_{0}^{\pi /6}{\cos x dx}}\)
3. \({\int_{0}^{\sqrt{3}/2}}\dfrac{1}{\sqrt{1-x^{2}}}\,{dx}\)
4. \({\int_{1}^{2}}\dfrac{1}{x} {dx}\)

Solution (a) An antiderivative of \(f(x) =x^{2}\) is \(F(x) =\dfrac{x^{3}}{3}\). By Part 2 of the Fundamental Theorem of Calculus, \[ {\int_{-2}^{1}{x^{2}}}dx=\left[ {\dfrac{{x^{3}}}{{3}}}\right] _{-2}^{1}={\dfrac{{1}^{3}}{{3}}}-{\dfrac{{\left( {-2}\right) ^{3}}}{{3}}}={\dfrac{{1}}{{3}}}+{\dfrac{{8}}{{3}}}={\dfrac{{9}}{{3}}}=3 \]

(b) An antiderivative of \(f(x) =\cos x\) is \(F(x) =\sin x\). By Part 2 of the Fundamental Theorem of Calculus, \[ \int_{0}^{\pi /6}\cos x\,dx= \Big[\sin x\Big] _{0}^{\pi /6}=\sin \dfrac{\pi }{6}-\sin 0=\dfrac{1}{2} \]

(c) An antiderivative of \(f(x) =\)\(\dfrac{1}{\sqrt{1-x^{2}}}\) is \(F(x) =\sin ^{-1}x\), provided \(\vert x\vert \lt1\). By Part 2 of the Fundamental Theorem of Calculus, \[ {\int_{0}^{\sqrt{3}/2}} \dfrac{1}{\sqrt{1-x^{2}}}\,dx= \Big[\sin^{-1}x\Big] _{0}^{\sqrt{3}/2}=\sin ^{-1}\dfrac{\sqrt{3}}{2}-\sin ^{-1}0=\dfrac{\pi}{3}-0=\dfrac{\pi}{3} \]

(d) An antiderivative of \(f(x)=\dfrac{1}{x}\) is \(F(x) =\ln \,\vert x\vert \). By Part 2 of the Fundamental Theorem of Calculus, \[ {\int_{1}^{2}} \dfrac{1}{x}\,dx= \Big[\ln \vert x\vert\Big] _{1}^{2}=\ln 2-\ln 1=\ln 2-0=\ln 2 \]