Suppose \(f\) is continuous on \([a,b]\).
- If \(g(x)=\int_{a}^{x}f(t)dt\), then \(g(x)\) is continuous on \([a,b]\) and differentiable on \((a,b)\) and \(g'(x)=f(x)\).
- \(\int_{a}^{b}f(x)dx=F(x)\Big|_{a}^{b}=F(b)-F(a)\), where \(F\) is any antiderivative of \(f\), that is, \(F'=f\).
Example
If \(g(x)=\int_{5}^{-x^{3}}e^{3t^{2}+1}dt\), find \(g'(x)\).
Solution
Let \(u=-x^3\). Then, \(\frac{du}{dx}=-3x^2\).
By Chain Rule, compute \(g'(x)\):
\[\begin{aligned}
g^{\prime}(x)& =\frac{d(g(x))}{dx}=\frac{d}{dx}\biggl(\int_{5}^{-x^{3}}e^{3t^{2}+1}dt\biggr) \\
&=\frac{d}{dx}\biggl(\int_{5}^{u}e^{3t^{2}+1}dt\biggr)=\frac{d}{du}\biggl(\int_{5}^{u}e^{3t^{2}+1}dt\biggr)\frac{du}{dx} \\
&=(e^{3u^2+1})(-3x^2) \\
&=-3x^{2}e^{3(-x^{3})^{2}+1}=-3x^{2}e^{3x^{6}+1}
\end{aligned}\]
Evaluate each integral.
- \(\int_1^2\Bigg(x^2+\frac1x\Bigg)dx\)
- \(\int_{2}^{4}\Bigl(x^{1/2}+e^{4x}\Bigr)dx\)
Solution
- \[\begin{aligned}
\int_{1}^{2}\left(x^{2}+\frac{1}{x}\right)dx& =\left(\frac{x^{3}}{3}+\ln x\right)\Bigg|_{1}^{2} \\
&=\left(\frac{2^{3}}{3}+\ln2\right)-\left(\frac{1^{3}}{3}+\ln1\right) \\
&=\frac73+\ln2
\end{aligned}\] - \[\begin{aligned}
\int_{4}^{9}\Bigl(x^{1/2}+e^{4x}\Bigr)dx& =\left(\frac{x^{3/2}}{\frac{3}{2}}+\frac{e^{4x}}{4}\right)\Bigg|_{4}^{9} \\
&=\left(\frac{2x^{3/2}}{3}+\frac{e^{4x}}{4}\right)\Bigg|_{4}^{9} \\
&=\left(\frac{2(9)^{3/2}}{3}+\frac{e^{36}}{4}\right)-\left(\frac{2(4)^{3/2}}{3}+\frac{e^{16}}{4}\right) \\
&=\frac{54}3+\frac{e^{36}}4-\frac{16}3-\frac{e^{16}}4 \\
&=\frac{38}3+\frac{e^{36}}4-\frac{e^{16}}4
\end{aligned}\]