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 \((a,b).\)

Using Part 1 of the Fundamental Theorem of Calculus

(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.

Using Part 1 of the Fundamental Theorem of Calculus

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.

Using Part 1 of the Fundamental Theorem of Calculus

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.

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

NOW WORK

Problems 25 and 31.

Finding the Area Under a Graph

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.

Interpreting an Integral Whose Integrand Is a Rate of Change

1. The function \(P=P(t) \) relates the population \(P\) (in billions of persons) as a function of the time \(t\) (in years). Suppose \(\int_{0}^{10}P^\prime (t)\, dt=3.\) Since \(P^{\prime} (t)\) is the rate of change of the population with respect to time, then the change in population from \(t=0\) to \(t=10\) is \(3\) billion persons.
2. The function \(v=v(t) \) is the speed \(v\) (in meters per second) of an object at a time \(t\) (in seconds). If \(\int_{0}^{5}v(t) dt=8,\) then the object travels 8 m during the interval \(0\leq t\leq 5.\) Do you see why? The speed \(v=v(t) \) is the rate of change of distance \(s\) with respect to time \(t.\) That is, \(v=\dfrac{ds}{dt}.\)

NOW WORK

Problem 51.