8.4 Derivatives of Integrals

In the previous section, we learned that the Fundamental Theorem of Calculus says that if $f$ is continuous on $[a,x]$, then \[\bbox[#F2F2F2,5px,border:2px solid black]{\frac{d}{d{\color{red}x}}\int_{a}^{{\color{red}x}}f({\color{blue}t})d{\color{blue}t}=f({\color{red}x})}\]

Example 1

Verify that \[\frac{d}{dx}\int_{1}^{x}f(t)dt=f(x)\] for $f(t)=t^{2}$.

Solution 1

First, let’s find $\int_{1}^{x}f(t)dt$: \[\begin{aligned} \int_{1}^{x}t^{2}dt & =\left.\frac{t^{3}}{3}\right]_{1}^{x}\\ & =\frac{x^{3}}{3}-\frac{1}{3}\end{aligned}\] Now we differentiate the result \[\frac{d}{dx}\left(\frac{x^{3}}{3}-\frac{1}{3}\right)=x^{2}=f(x).\] So the formula holds.

Example 2

Let ${\displaystyle F(x)=\int_{-1}^{x}\frac{1}{2+2t+t^{2}}dt.}$ Find $F'(3)$.

Solution 2

Because $t^{2}+2t+2=(t+1)^{2}+1$ is never zero, the integrand \[f(t)=\frac{1}{2+2t+t^{2}}\] is continuous everywhere (including on $[-1,3]$).

[Recall that a rational function $R(x)=\frac{P(x)}{Q(x)}$ where $P$ and $Q$ are two polynomials is continuous at $x=a$ if $Q(a)\neq0$.]

Therefore, we can apply the Fundamental Theorem of Calculus and conclude \[F'(3)=f(3)=\frac{1}{2+2(3)+3^{2}}=\frac{1}{17}.\]

We may use the chain rule in conjuction with the first part of the Fundamental Theorem of Calculus to find the derivative if the upper limit or the lower limit integration is a function. For example, if \[F(x)=\int_{a}^{g(x)}f(t)dt,\] to find $F'(x)$, let $u=g(x)$. Then
\[\begin{aligned}
\frac{dF}{dx} & =\frac{dF}{du}\frac{du}{dx}\\
& =\left(\frac{d}{du}\int_{a}^{u}f(t)dt\right)g'(x)\\
& =f(u)g'(x) &&{\small \text{ (by the Fundamental Theorem of Calculus)}}\\
& =f(g(x))g'(x) &&{\small (u=g(x))}.
\end{aligned}\]

Example 3

Find $S'(x)$, if \[S(x)=\int_{0}^{x^{2}}\frac{\sin t}{t}dt.\]

Solution 3

Let $u=x^{2}$. Then \[\begin{aligned} S'(x) & =\left(\frac{d}{du}\int_{0}^{u}\frac{\sin t}{t}dt\right)\frac{du}{dx}\\ & =\left(\frac{\sin u}{u}\right)(2x)\\ &\begin{equation*} =\frac{\sin x^{2}}{x^{2}}(2x)\tag{$u=x^{2}$}\end{equation*}\\ & =\frac{2\sin x^{2}}{x}\end{aligned}\]

Example 4

Find \[\frac{d}{dx}\left(\int_{x^{2}}^{x^{3}}\ln t\ dt\right),\qquad x>0.\]

Solution 4

We write \[\begin{aligned} \int_{x^{2}}^{x^{3}}\ln t\ dt & =\int_{x^{2}}^{c}\ln t\ dt+\int_{c}^{x^{3}}\ln t\ dt\\ & =-\int_{c}^{x^{2}}\ln t\ dt+\int_{c}^{x^{3}}\ln t\ dt,\end{aligned}\] where $c$ is an arbitrary constant. Because the derivative of a sum is the sum of derivatives, we have \[\frac{d}{dx}\left(\int_{x^{2}}^{x^{3}}\ln t\ dt\right)=-\frac{d}{dx}\left(\int_{c}^{x^{2}}\ln t\ dt\right)+\frac{d}{dx}\left(\int_{c}^{x^{3}}\ln t\ dt\right).\] Now we can differentiate with respect to $x$ like the previous example: \[\begin{aligned} \frac{d}{dx}\left(\int_{c}^{x^{2}}\ln t\ dt\right) &\begin{equation*} =\left(\frac{d}{du}\int_{c}^{u}\ln t\ dt\right)\frac{du}{dx}\tag{$u=x^{2}$}\\\end{equation*} & =(\ln u)(2x)\\ & =(\ln x^{2})(2x)\\ & =2x\ln x^{2}.\end{aligned}\] \[\begin{aligned} \frac{d}{dx}\left(\int_{c}^{x^{3}}\ln t\ dt\right) &\begin{equation*} =\left(\frac{d}{dv}\int_{c}^{v}\ln t\ dt\right)\frac{dv}{dx}\tag{$v=x^{3}$}\end{equation*}\\ & =(\ln v)(3x^{2})\\ & =3x^{2}\ln x^{3}.\end{aligned}\] Therefore, \[\frac{d}{dx}\left(\int_{x^{2}}^{x^{3}}\ln t\ dt\right)=-2x\ln x^{2}+3x^{2}\ln x^{3}\] We may further simplify the result using the fact that $\ln x^{r}=r\ln x$. So \[\begin{aligned} \frac{d}{dx}\left(\int_{x^{2}}^{x^{3}}\ln t\ dt\right) & =-4x\ln x+9x^{2}\ln x\\ & =x(9x-4)\ln x.\end{aligned}\]

Example 5

Find \[\lim_{x\to0^{+}}\frac{\int_{0}^{4x^{2}}\sin\sqrt{t}dt}{x^{3}}.\]

Solution 5

Because as $x\to0$, \[\int_{0}^{4x^{2}}\sin\sqrt{t}\ dt\to0,\quad x^{3}\to0,\] we have an indeterminate limit of type $0/0$. So we can apply l’Hôpital’s rule: \[\lim_{x\to0^{+}}\frac{\int_{0}^{4x^{2}}\sin\sqrt{t}dt}{x^{3}}\stackrel{H}{=}\lim_{x\to0^{+}}\frac{\frac{d}{dx}\int_{0}^{4x^{2}}\sin\sqrt{t}\ dt}{\frac{d}{dx}x^{3}}.\] Because \[\begin{aligned} \frac{d}{dx}\int_{0}^{4x^{2}}\sin\sqrt{t}\ dt & \begin{equation*}=\left(\frac{d}{du}\int_{0}^{u}\sin\sqrt{t}\ dt\right)\frac{du}{dx}\tag{$u=4x^{2}$}\end{equation*}\\ & =(\sin\sqrt{u})(8x)\\ & =8x\sin\sqrt{4x^{2}},\end{aligned}\] we get \[\begin{aligned} \lim_{x\to0^{+}}\frac{\frac{d}{dx}\int_{0}^{4x^{2}}\sin\sqrt{t}\ dt}{\frac{d}{dx}x^{3}} & =\lim_{x\to0^{+}}\frac{8x\sin\sqrt{4x^{2}}}{3x^{2}}\\ & =\lim_{x\to0^{+}}\frac{8x\sin2x}{3x^{2}}&&\small(\sqrt{4x^{2}}=|2x|=2x,\ x>0)\\ & =\lim_{x\to0^{+}}\frac{8}{3}\frac{\sin2x}{x}\\ & \stackrel{H}{=}\frac{8}{3}\lim_{x\to0^{+}}\frac{2\cos2x}{1}\\ & =\frac{16}{3}.\end{aligned}\]

Example 6

Find $dy/dx$ if $y$ is implicitly defined by the following equation: \[\int_{0}^{y}e^{-t^{2}}dt+\int_{0}^{x^{2}}\sin t\ dt=0.\]

Solution 6

Using implicit differentiation, we get \[\frac{d}{dx}\left(\int_{0}^{y}e^{-t^{2}}dt\right)+\frac{d}{dx}\left(\int_{0}^{x^{2}}\sin t\ dt\right)=0\] \[\frac{d}{dy}\left(\int_{0}^{y}e^{-t^{2}}dt\right)\frac{dy}{dx}+\frac{d}{dx}\left(\int_{0}^{x^{2}}\sin t\ dt\right)=0\]

By Part 1 of the Funamental Theorem of Calculus (FTC1):

\[e^{-y^{2}}\frac{dy}{dx}+\frac{d}{dx}\left(\int_{0}^{x^{2}}\sin t\ dt\right)=0 \]

\[e^{-y^{2}}\frac{dy}{dx}+\frac{d}{du}\left(\int_{0}^{u}\sin t\ dt\right)\frac{du}{dx}=0\tag{$u=x^{2}$}\] \[e^{-y^{2}}\frac{dy}{dx}+\sin u\,\frac{du}{dx}=0\tag{FTC1}\] \[e^{-y^{2}}\frac{dy}{dx}+\left(\sin x^{2}\right)(2x)=0\tag{$u=x^{2}$} \] \[\begin{aligned} \frac{dy}{dx} & =\frac{1}{e^{-y^{2}}}2x\ \sin x^{2}\\ & =2x\ e^{y^{2}}x^{2}.\end{aligned}\]