Short and Sweet Calculus

## 6.4 Derivatives of Integrals

In the previous section, we learned that the first part of the Fundamental Theorem of Calculus says that if $$f$$ is continuous on $$[a,x]$$, then $\boxed{\frac{d}{d\boldsymbol{x}}\int_{a}^{\boldsymbol{x}}f(\boldsymbol{t})d\boldsymbol{t}=f(\boldsymbol{x})}$

Example 6.7. Verify that $\frac{d}{dx}\int_{1}^{x}f(t)dt=f(x)$ for $$f(t)=t^{2}$$.

Solution

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

Solution

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 of an integral 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)\tag{\small by Fundamental Theorem of Calculus}\\ & =f(g(x))g'(x).\end{aligned}

Example 6.9. Find $$S'(x)$$, if $S(x)=\int_{0}^{x^{2}}\frac{\sin t}{t}dt.$

Solution

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)\\ & =\frac{\sin x^{2}}{x^{2}}(2x)\tag{\ensuremath{u=x^{2}}}\\ & =\frac{2\sin x^{2}}{x}.\end{aligned}