OBJECTIVES

When you finish this section, you should be able to:

1. Use Part 1 of the Fundamental Theorem of Calculus (p. 363)
2. Use Part 2 of the Fundamental Theorem of Calculus (p. 365)
3. Interpret an integral using Part 2 of the Fundamental Theorem of Calculus (p. 365)

NEED TO REVIEW?

Antiderivatives are discussed in Section 4.8, pp. 328-331.

THEOREM Fundamental Theorem of Calculus, Part 1

Let $f$ be a function that is continuous on a closed interval $[a,b]$. The function $I$ defined by $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[5pt]{ I\,(x)=\int_{a}^{x}f(t)\,dt } }$$

has the properties that it is continuous on $[a,b]$ and differentiable on $(a,b).$ Moreover, $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[5pt]{ I' (x)=\dfrac{d}{dx}\left[\int_{a}^{x}f(t)\,dt \right] =f(x) } }$$

for all $x$ in $(a,b).$

(a) $\dfrac{d}{dx}\int_{0}^{x}\sqrt{t+1}\,dt=\sqrt{ x+1}$

(b) $\dfrac{d}{dx}\int_{2}^{x}\dfrac{s^{3}-1}{2s^{2}+s+1}ds=\dfrac{x^{3}-1}{2x^{2}+x+1}$

NOW WORK

Problem 5.

Find $\dfrac{d}{dx}\int_{4}^{3x^{2}+1}\sqrt{e^{t}+t}\,dt$.

Solution The upper limit of integration is a function of $x$, so we use the Chain Rule along with Part 1 of the Fundamental Theorem of Calculus.

Let $y=\int_{4}^{3x^{2}+1}\sqrt{e^{t}+t}\,dt$ and $u(x) =3x^{2}+1.$ Then $y=\int_{4}^{u}\sqrt{e^{t}+t}\,dt$ and $$\begin{eqnarray*} \dfrac{d}{dx}\int_{4}^{3x^{2}+1}\sqrt{e^{t}+t}\,dt &=& \dfrac{dy}{dx} \underset{\underset{{\color{#0066A7}{\hbox{Chain Rule}}}}{\color{#0066A7}{\uparrow}}}{=} \dfrac{dy}{du}\cdot \dfrac{du}{dx}= \left[ \dfrac{d}{du}\int_{4}^{u}\sqrt{e^{t}+t}\,dt\right] \cdot \dfrac{du}{dx} \\ & \underset{\underset{\color{#0066A7}{{\hbox{Use the Fundamental Theorem}}}} {\color{#0066A7}{\uparrow}}}{=}& \sqrt{e^{u}+u}\cdot \dfrac{du}{dx} \underset{\underset{{\color{#0066A7}{{\hbox{\({u=3x}^{2}+1\); \(\dfrac{du}{dx}=6x\)}}}}{\hbox{\( \)}}}{\color{#0066A7}{\uparrow}}}{=}\sqrt{e^{( 3x^{2}+1) }+3x^{2}+1}\cdot 6x \\ \end{eqnarray*}$$

NOW WORK

Problem 11.

Find $\dfrac{d}{dx}\int_{x^{3}}^{5}(t^{4}+1)^{1/3}\,dt$.

Solution To use Part 1 of the Fundamental Theorem of Calculus, the variable must be part of the upper limit of integration. So, we use the fact that $\int_{a}^{b}f(x)\,dx=-\int_{b}^{a}f(x)\,dx$ to interchange the limits of integration. $$\dfrac{d}{dx}\int_{x^{3}}^{5}(t^{4}+1)^{1/3}\,dt=\dfrac{d}{dx}\left[ {-} \int_{5}^{x^{3}}(t^{4}+1)^{1/3} dt\right] ={-}\dfrac{d}{dx} \int_{5}^{x^{3}}(t^{4}+1)^{1/3} dt$$

Now we use the Chain Rule. We let $y=\int_{5}^{x^{3}}( t^{4}+1)^{1/3}dt$ and $u(x) =x^{3}.$

$$\begin{eqnarray*} \dfrac{d}{dx}\int_{x^{3}}^{5}(t^{4}+1)^{1/3}\,dt &=& -\dfrac{d}{dx}\int_{5}^{x^{3}}(t^{4}+1)^{1/3}\,dt=-\dfrac{dy}{dx} \underset{ \underset{ \color{#0066A7}{\hbox{Chain Rule}} } {\color{#0066A7}{\uparrow}} } {=} -\dfrac{dy}{du}\cdot\dfrac{du}{dx}\\ &=&-\dfrac{d}{du}\int_{5}^{u}( t^{4}+1) ^{1/3}dt\cdot \dfrac{du}{dx} \\ &=& -(u^{4}+1)^{1/3}\cdot \dfrac{du}{dx} {\color{#0066A7}\enspace{\color{#0066A7}{\hbox{Use the Fundamental Theorem}}}} \\ &=&-( x^{12}+1) ^{1/3}\cdot 3x^{2} \enspace{\color{#0066A7}{{\hbox{\({u=x}^{3}\); \(\dfrac{du}{dx}=3x^{2}\)}}}} \\ &=&-3x^{2}(x^{12}+1)^{1/3} \end{eqnarray*}$$

NOW WORK

Problem 15.

THEOREM Fundamental Theorem of Calculus, Part 2

Let $f$ be a function that is continuous on a closed interval $[a,b]$. If $F$ is any antiderivative of $f$ on $[a,b]$, then $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[5pt]{ \int_{a}^{b}f(x)\,dx=F(b)-F(a)}}$$

Proof

Let $I(x) =\int_{a}^{x}f(t)\, dt.$ Then from Part 1 of the Fundamental Theorem of Calculus, $I=I(x)$ is continuous for $a\leq x\leq b$ and differentiable for $a\lt x\lt b$. So, $$\dfrac{d}{dx}\int_{a}^{x}f(t) ~dt=f(x) \qquad a\lt x\lt b$$

That is, $\int_{a}^{x}f(t)\,dt$ is an antiderivative of $f$. So, if $F$ is any antiderivative of $f$, then $$F(x)=\int_{a}^{x}f(t)\,dt+C$$

where $C$ is some constant. Since $F$ is continuous on $[a,b] ,$ we have

$$F(a)=\int_{a}^{a}f(t) dt+C \qquad F(b)=\int_{a}^{b}f(t) dt+C$$

Since, ${\int_{a}^{a}{f(t) dt}=0}$, subtracting $F(a)$ from $F(b)$ gives $$F(b)-F(a)=\int_{a}^{b}f(t) dt$$

Since $t$ is a dummy variable, we can replace $t$ by $x$ and the result follows.

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

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, $${\int_{1}^{2}} \dfrac{1}{x}\,dx= \Big[\ln \vert x\vert\Big] _{1}^{2}=\ln 2-\ln 1=\ln 2-0=\ln 2$$

NOW WORK

Problems 25 and 31.

Find the area under the graph of $f(x) =e^{x}$ from $-1$ to $1.$

Solution Figure 23 shows the graph of $f(x)=e^{x}$ on the closed interval $[-1,1]$.

The area $A$ under the graph of $f$ from $-1$ to $1$ is given by $$A= \int\nolimits_{-1}^{1}e^{x}\, dx=\Big[e^{x}\Big] _{-1}^{1}=e^{1}-e^{-1} = e-\dfrac{1}{e} \approx 2.35.$$

NOW WORK

Problem 45.

NOW WORK

Problem 51.