The Relation Between Integration and Differentiation

Differentiation and integration seem to answer two unrelated questions. The derivative gives the slope of a curve at a point; the definite integral gives the area under a curve. Yet the two operations undo each other, much as squaring and taking the square root undo each other for positive numbers. This connection is the content of the two fundamental theorems of calculus, and it is what makes most integrals computable: instead of a limit of sums, we only need a function whose derivative is the integrand.

Here is a first example. Take ${f(t)=t^2}$ and integrate it from $0$ to a variable upper limit $x.$ From the definition of the integral as a limit of sums (see Definite Integrals) one finds

$$\int_0^x t^2\,dt=\frac{x^3}{3}.$$

The result is a new function of $x,$ and its derivative is ${\left(\frac{x^3}{3}\right)'=x^2},$ exactly the function we started with. Integrating $f$ and then differentiating gives $f$ back. The rest of this page shows that this is no accident.

On this page: The area function · First fundamental theorem · Zero-derivative theorem · Primitive functions · Second fundamental theorem · Examples · Properties from the derivative · Exercises

The Area Function

Let $f$ be integrable on an interval $[a,b]$ and fix a point $c$ in $[a,b].$ For every $x$ in $[a,b]$ put

$$A(x)=\int_c^x f(t)\,dt.$$

When ${f\ge0}$ and ${x\gt c},$ the number $A(x)$ is the area of the region under the graph of $f$ between $c$ and $x.$ As $x$ moves, the area changes, so $A$ is a function of the upper limit. It is called the integral with a variable upper limit.

• The letter $t$ is only the variable of integration. Any other letter would do, but not $x,$ which is already the upper limit.

• ${A(c)=0},$ and for ${x\lt c}$ the integral means ${-\int_x^c f(t)\,dt}.$

• Another starting point changes $A$ only by a constant: ${\int_{c_1}^x f(t)\,dt}={\int_{c_1}^{c_2}f(t)\,dt}+{\int_{c_2}^x f(t)\,dt},$ and the first integral on the right does not depend on $x.$

The First Fundamental Theorem of Calculus

In other words, ${\frac{d}{dx}\int_c^x f(t)\,dt=f(x)}\text{:}$ the derivative of an integral with respect to its upper limit is the integrand at that limit. If $f$ is continuous on the whole interval, this holds at every inner point. The same works on an open interval $I,$ because every point of $I$ is an inner point of some ${[a,b]\subset I}$ that also contains $c.$ So every function continuous on an open interval is the derivative of some function there.

Why the Theorem Is True

Take a small ${h\gt0}.$ The difference ${A(x+h)-A(x)}$ is the area from $c$ to ${x+h}$ minus the area from $c$ to $x,$ that is, the area of the thin strip between $x$ and ${x+h}\text{:}$

$$A(x+h)-A(x)=\int_x^{x+h}f(t)\,dt.$$

The strip has width $h,$ and when $h$ is small, its height stays close to $f(x).$ So the strip is almost a rectangle of area ${h\,f(x)},$ and

$$\frac{A(x+h)-A(x)}{h}\approx f(x).$$

The smaller $h,$ the better the approximation. In the limit, the difference quotient on the left becomes the derivative $A'(x),$ and the approximation becomes the equality ${A'(x)=f(x)}.$

If $f$ is continuous on ${[x,x+h]},$ the mean value theorem for integrals makes this exact: the strip has the area ${h\,f(z)}$ for some point $z$ between $x$ and ${x+h}.$ When ${h\to0},$ the point $z$ is squeezed to $x,$ and ${f(z)\to f(x)}$ because $f$ is continuous.

Proof

The picture assumes that $f$ is continuous near $x,$ while the theorem asks for continuity at the single point $x$ only. Here is a proof that uses nothing more.

Fix a point $x$ of $(a,b)$ where $f$ is continuous, and let ${h\ne0}$ be so small that ${x+h}$ is still in $[a,b].$ As above, ${A(x+h)-A(x)}={\int_x^{x+h}f(t)\,dt}.$ The number $f(x)$ is a constant, so ${\int_x^{x+h}f(x)\,dt=h\,f(x)},$ and therefore

${\frac{A(x+h)-A(x)}{h}}-{f(x)}={{\frac{1}{h}\int_x^{x+h}\big(f(t)-f(x)\big)\,dt}.}$

We show that the right-hand side tends to $0.$ Let ${\varepsilon\gt0}.$ Because $f$ is continuous at $x,$ there is a ${\delta\gt0}$ such that ${\lvert f(t)-f(x)\rvert\lt\varepsilon}$ for every $t$ in $[a,b]$ with ${\lvert t-x\rvert\lt\delta}.$ Now take ${0\lt\lvert h\rvert\lt\delta}.$ Every $t$ between $x$ and ${x+h}$ is closer to $x$ than $\delta,$ so the integrand is less than $\varepsilon$ in absolute value on an interval of length $\lvert h\rvert,$ and

${\left\lvert\frac{1}{h}\int_x^{x+h}\big(f(t)-f(x)\big)\,dt\right\rvert}\le{\frac{1}{\lvert h\rvert}\cdot\varepsilon\,\lvert h\rvert}={{\varepsilon}.}$

(For ${h\lt0}$ the integral runs from right to left, which changes only its sign, so the estimate is the same.) Thus the difference quotient differs from $f(x)$ by at most $\varepsilon$ whenever ${0\lt\lvert h\rvert\lt\delta}.$ Since $\varepsilon$ was arbitrary,

${A'(x)}={\lim_{h\to0}\frac{A(x+h)-A(x)}{h}}={{f(x)}.}$

Examples

Example 1. ${\frac{d}{dx}\int_1^x(t^3+1)\,dt}={x^3+1}.$ The lower limit plays no part: the derivative is the integrand with $t$ replaced by $x.$ We can check it: by the second fundamental theorem below, ${\int_1^x(t^3+1)\,dt}={\frac{x^4}{4}+x-\frac{5}{4}},$ and the derivative of this is ${x^3+1}.$

Example 2. ${\frac{d}{dx}\int_0^x\sin(t^2)\,dt=\sin(x^2)}.$ The integral ${\int_0^x\sin(t^2)\,dt}$ cannot be written with the elementary functions (powers, roots, exponentials, logarithms, trigonometric and inverse trigonometric functions and their combinations), yet its derivative is simple. In fact, integrals with a variable upper limit are a way to define new functions, and the theorem gives their derivatives at once. For example, the natural logarithm can be defined as ${\ln x=\int_1^x\frac{dt}{t}}$ for ${x\gt0},$ and then ${(\ln x)'=\frac{1}{x}}$ at once.

Example 3 (a variable lower limit). Swapping the limits changes the sign of an integral: ${\int_x^2e^{t^2}\,dt=-\int_2^xe^{t^2}\,dt}.$ Therefore

$$\frac{d}{dx}\int_x^2e^{t^2}\,dt=-e^{x^2}.$$

Example 4 (a function in the upper limit). Find the derivative of ${F(x)=\int_0^{x^2}\sqrt{1+t^3}\,dt}.$

Let ${G(u)=\int_0^u\sqrt{1+t^3}\,dt}.$ Then ${F(x)=G(x^2)}$ and ${G'(u)=\sqrt{1+u^3}},$ so by the chain rule

${F'(x)}={G'(x^2)\cdot(x^2)'}={{2x\sqrt{1+x^6}}.}$

Both cases together give the general rule. If $f$ is continuous on an interval that contains all the values of $u$ and $v,$ and $u$ and $v$ are differentiable, then

To see this, split the integral at a fixed point $c$ and apply Examples 3 and 4 to the two parts.

Example 5. Here ${u(x)=2x}$ and ${v(x)=x^2}\text{:}$

${\frac{d}{dx}\int_{2x}^{x^2}\frac{dt}{1+t^2}}={\frac{1}{1+x^4}\cdot2x}-{\frac{1}{1+4x^2}\cdot2}={{\frac{2x}{1+x^4}-\frac{2}{1+4x^2}}.}$

When the Integrand Has a Jump

The continuity of $f$ at the point $x$ cannot be dropped. Let ${f(t)=1}$ for ${t\lt0}$ and ${f(t)=2}$ for ${t\ge0},$ and let ${A(x)=\int_0^x f(t)\,dt}.$ Then ${A(x)=x}$ for ${x\le0}$ and ${A(x)=2x}$ for ${x\ge0}.$

The function $A$ is continuous everywhere, and ${A'(x)=f(x)}$ at every ${x\ne0}.$ At ${x=0},$ where $f$ jumps, the graph of $A$ has a corner: its slope is $1$ on the left and $2$ on the right, so $A'(0)$ does not exist. Integration makes a function smoother: where $f$ jumps, $A$ only has a corner.

The Zero-Derivative Theorem

If a function is constant on an interval, its derivative is $0$ there, because every difference quotient ${\frac{f(x+h)-f(x)}{h}}$ is $0.$ The converse is also true, and it is the key to the second fundamental theorem.

Proof. Take any two points ${x_1\lt x_2}$ of $I.$ The function $f$ is differentiable, and hence continuous, on ${[x_1,x_2]}.$ By the mean value theorem (Lagrange's theorem) there is a point $\xi$ between $x_1$ and $x_2$ such that

${f(x_2)-f(x_1)}={f'(\xi)\,(x_2-x_1)}={{0}.}$

So ${f(x_1)=f(x_2)}$ for any two points of $I,$ which means that $f$ is constant.

The word interval matters. The function ${g(x)=\frac{\lvert x\rvert}{x}}$ is defined for ${x\ne0},$ and ${g'(x)=0}$ wherever it is defined, but $g$ is not constant: it equals $-1$ for ${x\lt0}$ and $1$ for ${x\gt0}.$ Its domain consists of two intervals, and the theorem only says that $g$ is constant on each of them.

Primitive Functions and the Second Fundamental Theorem of Calculus

Definition of a Primitive Function

For example, $\sin x$ is a primitive of $\cos x$ on the whole real line, because ${(\sin x)'=\cos x}.$ The function ${\frac{x^3}{3}}$ is a primitive of $x^2,$ and so are ${\frac{x^3}{3}+5}$ and ${\frac{x^3}{3}-\sqrt{2}}\text{:}$ adding a constant does not change the derivative. That is why we say a primitive and not the primitive. Adding a constant is also the only freedom there is:

Proof. ${(P+C)'=P'=f},$ so every function ${P+C}$ is a primitive of $f.$ Conversely, if $Q$ is any primitive of $f$ on $I,$ then ${(Q-P)'=f-f=0}$ on $I,$ and by the zero-derivative theorem ${Q-P}$ is a constant $C.$

Geometrically, the graphs of the primitives of $f$ are copies of one graph shifted up and down. Above each point $x$ all of them have the same slope $f(x).$ In the picture, ${f(x)=x^2-1},$ and every curve has the slope ${f(1.5)=1.25}$ at ${x=1.5}.$

The family of all primitives of $f$ is written with the integral sign without limits, ${\int f(x)\,dx=P(x)+C},$ and is called the indefinite integral of $f.$ There is more about this notation in Integrals - 2, and a long list of formulas in Indefinite Integrals.

Does every function have a primitive? Every continuous function does: by the first fundamental theorem, ${A(x)=\int_c^x f(t)\,dt}$ is one. But not every function has a primitive. The function with a jump from the example above has none on the whole line. By the zero-derivative theorem, a primitive would be ${x+C_1}$ for ${x\lt0}$ and ${2x+C_2}$ for ${x\gt0}.$ It must be continuous at $0,$ so ${C_1=C_2},$ and then it has a corner at $0$ and no derivative there.

The Second Fundamental Theorem of Calculus

Proof. Let ${A(x)=\int_c^x f(t)\,dt}.$ Since $f$ is continuous at every point of $I,$ the first fundamental theorem gives ${A'(x)=f(x)}$ on $I,$ so $A$ is a primitive of $f.$ Two primitives of the same function differ by a constant, so ${A(x)=P(x)+k}$ for some number $k.$ At ${x=c}$ this gives ${0=A(c)=P(c)+k},$ so ${k=-P(c)}$ and ${A(x)=P(x)-P(c)}.$

With ${c=a}$ and ${x=b}$ this is the formula used to compute definite integrals, the Newton–Leibniz formula. If $f$ is continuous on an open interval that contains ${[a,b]}$ and $P$ is a primitive of $f$ there, then

The bracket ${\Bigl[P(x)\Bigr]_a^b},$ also written ${P(x)\Big\vert_a^b},$ is short for ${P(b)-P(a)}.$ Any primitive gives the same result, since the constant cancels: ${\big(P(b)+C\big)}-{\big(P(a)+C\big)}={P(b)-P(a)}.$

The theorem turns a hard problem into a much easier one. Instead of a limit of sums, we need a function with a given derivative, and every differentiation formula read from right to left supplies one.

Continuity matters. The integrand must be continuous on the whole interval of integration. For instance, ${\left(-\frac{1}{x}\right)'=\frac{1}{x^2}},$ but the formula would give ${\int_{-1}^1\frac{dx}{x^2}}={\Bigl[-\frac{1}{x}\Bigr]_{-1}^1}={-2},$ a negative number as the integral of a positive function. The formula does not apply here: ${\frac{1}{x^2}}$ is not even defined at ${x=0},$ which lies in ${[-1,1]}.$ (This is an improper integral, and it diverges.)

Worked Examples

Example 1. The function ${\frac{x^3}{3}}$ is a primitive of $x^2,$ so

${\int_0^1x^2\,dx}={\Bigl[\frac{x^3}{3}\Bigr]_0^1}={\frac{1}{3}-0}={{\frac{1}{3}}.}$

Example 2. ${x^3-2x^2+5x}$ is a primitive of ${3x^2-4x+5},$ so

${\int_1^2(3x^2-4x+5)\,dx}={\Bigl[x^3-2x^2+5x\Bigr]_1^2}={(8-8+10)}-{(1-2+5)}={{6}.}$

Example 3. Write the root as a power with a negative exponent:

${\int_1^4\frac{dx}{\sqrt{x}}}={\int_1^4x^{-1/2}\,dx}={\Bigl[2\sqrt{x}\Bigr]_1^4}={4-2}={{2}.}$

Example 4. Since ${(-\cos x)'=\sin x},$

${\int_0^\pi\sin x\,dx}={\Bigl[-\cos x\Bigr]_0^\pi}={-\cos\pi+\cos0}={1+1}={{2}.}$

The area under one arch of the sine curve is exactly $2.$

Example 5. Since ${(\arctan x)'=\frac{1}{1+x^2}},$

${\int_0^1\frac{dx}{1+x^2}}={\Bigl[\arctan x\Bigr]_0^1}={\frac{\pi}{4}-0}={{\frac{\pi}{4}}.}$

Example 6. Since ${(\ln x)'=\frac{1}{x}}$ for ${x\gt0},$

${\int_1^e\frac{dx}{x}}={\Bigl[\ln x\Bigr]_1^e}={1-0}={{1}.}$

Example 7 (an integrand given by two formulas). Find ${\int_{-1}^2\lvert x\rvert\,dx}.$ For ${x\lt0}$ we have ${\lvert x\rvert=-x},$ and for ${x\ge0}$ we have ${\lvert x\rvert=x},$ so we split the integral at $0\text{:}$

${\int_{-1}^0(-x)\,dx+\int_0^2x\,dx}={\Bigl[-\frac{x^2}{2}\Bigr]_{-1}^0}+{\Bigl[\frac{x^2}{2}\Bigr]_0^2}={\frac{1}{2}+2}={{\frac{5}{2}}.}$

Another way: ${P(x)=\frac{x\lvert x\rvert}{2}}$ is a primitive of ${\lvert x\rvert}$ on the whole line (check the derivative separately for ${x\gt0},$ for ${x\lt0}$ and, from the definition, at ${x=0}$), and ${P(2)-P(-1)}={2-\left(-\frac{1}{2}\right)}={\frac{5}{2}}.$

Basic Primitives

Each row of this table is a derivative formula read backwards. Check any row by differentiating the right column.

Function $f(x)$	A primitive $P(x)$
$x^n,$ ${n\ne-1}$	$\frac{x^{n+1}}{n+1}$
$\frac{1}{x}$	$\ln\lvert x\rvert$
$e^x$	$e^x$
$\sin x$	$-\cos x$
$\cos x$	$\sin x$
$\frac{1}{\cos^2x}$	$\tan x$
$\frac{1}{1+x^2}$	$\arctan x$
$\frac{1}{\sqrt{1-x^2}}$	$\arcsin x$

Each formula holds on every open interval where $f$ is continuous: for example, ${\frac{1}{x}}$ on ${(0,\infty)}$ or on ${(-\infty,0)},$ but not on an interval that contains $0.$ For a non-integer $n,$ the power $x^n$ is taken for ${x\gt0}.$ So for ${0\lt a\lt b}$

${\int_a^bx^n\,dx=\frac{b^{n+1}-a^{n+1}}{n+1}}$  for every ${n\ne-1},$   and   ${\int_a^b\frac{dx}{x}=\ln\frac{b}{a}}.$

Properties of a Function Deduced from Its Derivative

If $f$ has a continuous derivative on an open interval $I,$ then $f$ is a primitive of $f',$ and the second fundamental theorem gives, for all $c$ and $x$ in $I,$

$$f(x)=f(c)+\int_c^x f'(t)\,dt.$$

The value of the function at $x$ is its value at $c$ plus the change accumulated between $c$ and $x.$ In physics: if $s(t)$ is the position of a moving body and ${v(t)=s'(t)}$ is its velocity, then ${s(t)}={s(0)+\int_0^t v(\tau)\,d\tau},$ so the displacement is the integral of the velocity. The formula lets us read properties of $f$ from properties of $f'.$

1. The sign of the derivative. If ${f'(t)\ge0}$ on $I,$ then for ${x\gt c}$ the integral ${\int_c^x f'(t)\,dt}$ is not negative, so ${f(x)\ge f(c)}\text{:}$ the function is increasing (never decreases) on $I.$ If ${f'(t)\gt0}$ on $I,$ the integral is positive, and $f$ is strictly increasing. In the same way, ${f'\le0}$ makes $f$ decreasing.

For example, ${f(x)=x+\sin x}$ has ${f'(x)=1+\cos x\ge0},$ so it is increasing on the whole real line. It is even strictly increasing. If ${f(x)=f(c)}$ for some ${x\gt c},$ the increasing function $f$ would be constant on ${[c,x]},$ and then ${f'=0}$ on all of ${[c,x]}.$ But ${1+\cos t=0}$ only at the isolated points ${t=\pi+2k\pi}.$

2. A bound for the derivative bounds the change. If ${\lvert f'(t)\rvert\le M}$ on $I,$ then

${\lvert f(x)-f(c)\rvert}={\left\lvert\int_c^x f'(t)\,dt\right\rvert}\le{{M\,\lvert x-c\rvert}.}$

For example, ${\lvert\cos t\rvert\le1}$ gives ${\lvert\sin x-\sin c\rvert\le\lvert x-c\rvert}$ for all real $x$ and $c.$

3. An increasing derivative makes the function convex. If $f'$ is increasing on $I,$ then $f$ is convex on $I\text{:}$ its graph bends upward. Take three points ${x_1\lt x_2\lt x_3}$ of $I.$ The slope of the chord over ${[x_1,x_2]}$ is

${\frac{f(x_2)-f(x_1)}{x_2-x_1}}={{\frac{1}{x_2-x_1}\int_{x_1}^{x_2}f'(t)\,dt},}$

the average value of $f'$ on ${[x_1,x_2]}.$ Because $f'$ is increasing, this average is at most ${f'(x_2)},$ while the average over ${[x_2,x_3]}$ is at least ${f'(x_2)}.$ So the chord over ${[x_1,x_2]}$ is never steeper than the chord over ${[x_2,x_3]}.$ This means that the point ${(x_2,f(x_2))}$ lies on or below the chord joining ${(x_1,f(x_1))}$ and ${(x_3,f(x_3))},$ which is the definition of convexity. In the same way, a decreasing $f'$ makes $f$ concave. For example, the derivative of $e^x$ is $e^x,$ which is increasing, so $e^x$ is convex.

Exercises

1. Find a primitive of ${f(x)=6x^2-4x+3}$ and use it to evaluate ${\int_0^2f(x)\,dx}.$

${P(x)=2x^3-2x^2+3x}$ is a primitive, since ${P'(x)=6x^2-4x+3}.$ Then ${\int_0^2f(x)\,dx}={P(2)-P(0)}={16-8+6}={{14}.}$ Answer: $14.$

2. Evaluate ${\int_1^8\sqrt[3]{x}\,dx}.$

${\sqrt[3]{x}=x^{1/3}},$ and a primitive is ${\frac{3}{4}x^{4/3}}.$ Since ${8^{4/3}=16},$ the integral is ${\frac{3}{4}(16-1)}={\frac{45}{4}}.$ Answer: $\frac{45}{4}.$

3. Evaluate ${\int_1^4\left(1+\sqrt{x}\right)^2dx}.$

${\left(1+\sqrt{x}\right)^2=1+2x^{1/2}+x},$ and a primitive is ${x+\frac{4}{3}x^{3/2}+\frac{x^2}{2}}.$ Its value at $4$ is ${4+\frac{32}{3}+8=\frac{68}{3}},$ at $1$ it is ${1+\frac{4}{3}+\frac{1}{2}=\frac{17}{6}},$ and ${\frac{68}{3}-\frac{17}{6}=\frac{119}{6}}.$ Answer: $\frac{119}{6}.$

4. Evaluate ${\int_1^2\frac{x^3+x-3}{x^3}\,dx}.$

Divide term by term: ${\frac{x^3+x-3}{x^3}}={1+x^{-2}-3x^{-3}}.$ A primitive is ${x-\frac{1}{x}+\frac{3}{2x^2}}.$ Its value at $2$ is ${2-\frac{1}{2}+\frac{3}{8}=\frac{15}{8}},$ at $1$ it is ${\frac{3}{2}},$ and ${\frac{15}{8}-\frac{3}{2}=\frac{3}{8}}.$ Answer: $\frac{3}{8}.$

5. Evaluate ${\int_0^{\pi/2}(3\sin x+2\cos x)\,dx}.$

${\Bigl[-3\cos x+2\sin x\Bigr]_0^{\pi/2}}={(0+2)-(-3+0)}={{5}.}$ Answer: $5.$

6. Evaluate ${\int_0^{\pi/4}\frac{dx}{\cos^2x}}.$

${\Bigl[\tan x\Bigr]_0^{\pi/4}}={1-0}={1}.$ Answer: $1.$

7. Evaluate ${\int_{-1}^3\lvert x-1\rvert\,dx}.$

Split at ${x=1}\text{:}$ ${\int_{-1}^1(1-x)\,dx}={\Bigl[x-\frac{x^2}{2}\Bigr]_{-1}^1}={\frac{1}{2}-\left(-\frac{3}{2}\right)}={2}$ and ${\int_1^3(x-1)\,dx}={\Bigl[\frac{x^2}{2}-x\Bigr]_1^3}={\frac{3}{2}-\left(-\frac{1}{2}\right)}={2}.$ Answer: $4$ (the area of two triangles with base $2$ and height $2$).

8. Find $F'(x)$ if ${F(x)=\int_2^x\sqrt{1+t^4}\,dt}.$

By the first fundamental theorem, ${F'(x)=\sqrt{1+x^4}}.$

9. Find $F'(x)$ if ${F(x)=\int_0^{x^3}\cos(t^2)\,dt}.$

By the chain rule, ${F'(x)}={\cos\big((x^3)^2\big)\cdot3x^2}={3x^2\cos(x^6)}.$

10. Find $F'(x)$ if ${F(x)=\int_x^{3x}e^{-t^2}\,dt}.$

By the rule for variable limits, ${F'(x)}={e^{-9x^2}\cdot3-e^{-x^2}\cdot1}={3e^{-9x^2}-e^{-x^2}}.$

11. Find a continuous function $f$ and all constants $c$ such that ${\int_c^x f(t)\,dt=x^2-4}$ for all $x.$

Differentiating both sides gives ${f(x)=2x}.$ Putting ${x=c}$ makes the left side $0,$ so ${c^2-4=0}.$ Answer: ${f(x)=2x},$ and ${c=2}$ or ${c=-2}.$

12. A continuous function $f$ satisfies ${\int_0^x f(t)\,dt=x^2\sin\pi x}$ for all $x.$ Find ${f\left(\frac{1}{2}\right)}$ and ${f'\left(\frac{1}{2}\right)}.$

Differentiate: ${f(x)}={{2x\sin\pi x+\pi x^2\cos\pi x},}$ so ${f\left(\frac{1}{2}\right)=1+0=1}.$ Differentiate again: ${f'(x)}={2\sin\pi x}+{4\pi x\cos\pi x-\pi^2x^2\sin\pi x},$ so ${f'\left(\frac{1}{2}\right)=2-\frac{\pi^2}{4}}.$

13. A continuous function $f$ satisfies ${\int_0^{f(x)}t^2\,dt=x^3+7x}$ for all ${x\ge0}.$ Find $f(1).$

The left side equals ${\frac{f(x)^3}{3}},$ so ${f(x)^3=3x^3+21x}.$ For ${x=1}\text{:}$ ${f(1)^3=24}.$ Answer: ${f(1)=\sqrt[3]{24}=2\sqrt[3]{3}}.$

14. A particle moves along a line with velocity ${v(t)=t^2-4t+3}$ (in m/s). Find its displacement from ${t=0}$ to ${t=3}$ and the total distance it travels in this time.

The displacement is ${\int_0^3v(t)\,dt}={\Bigl[\frac{t^3}{3}-2t^2+3t\Bigr]_0^3}={9-18+9}={{0}\text{:}}$ the particle returns to its starting point. Since ${v(t)=(t-1)(t-3)},$ it moves forward for ${0\lt t\lt1}$ and backward for ${1\lt t\lt3}\text{:}$ ${\int_0^1v(t)\,dt=\frac{4}{3}}$ and ${\int_1^3v(t)\,dt=-\frac{4}{3}}.$ Answer: the displacement is $0,$ and the distance is ${\frac{4}{3}+\frac{4}{3}=\frac{8}{3}}$ m.

15. Let ${f(x)=2+\int_0^x\frac{e^t}{1+t^2}\,dt}.$ Without evaluating the integral, find the quadratic polynomial ${p(x)=a+bx+cx^2}$ such that ${p(0)=f(0)},$ ${p'(0)=f'(0)}$ and ${p''(0)=f''(0)}.$

${f(0)=2}.$ ${f'(x)=\frac{e^x}{1+x^2}},$ so ${f'(0)=1}.$ ${f''(x)}={\frac{e^x(1+x^2)-e^x\cdot2x}{(1+x^2)^2}}={\frac{e^x(x-1)^2}{(1+x^2)^2}},$ so ${f''(0)=1}.$ Since ${p(0)=a},$ ${p'(0)=b}$ and ${p''(0)=2c},$ we get ${a=2},$ ${b=1},$ ${c=\frac{1}{2}}.$ Answer: ${p(x)=2+x+\frac{x^2}{2}}.$

16. Prove that ${\ln(1+x)\le x}$ for all ${x\ge0}.$

Let ${g(x)=x-\ln(1+x)}.$ Then ${g(0)=0}$ and ${g'(t)}={1-\frac{1}{1+t}}={\frac{t}{1+t}}\ge{0}$ for ${t\ge0},$ so ${g(x)}={g(0)+\int_0^xg'(t)\,dt}\ge{0}$ for ${x\ge0},$ which is the inequality.

17. Prove that no polynomial is a primitive of ${\frac{1}{x}}$ on ${(0,\infty)}.$

Suppose ${p'(x)=\frac{1}{x}}$ for all ${x\gt0}$ for some polynomial $p.$ Then ${x\,p'(x)-1=0}$ for all ${x\gt0}.$ The left side is a polynomial with infinitely many roots, so it is the zero polynomial, and ${x\,p'(x)=1}$ for every $x.$ But at ${x=0}$ the left side is $0,$ a contradiction.

18. Is there a function $f$ with a continuous second derivative such that ${f''(x)\gt0}$ for all $x,$ ${f'(0)=1}$ and ${f'(1)=0}?$

No. By the second fundamental theorem applied to $f',$ ${f'(1)-f'(0)}={\int_0^1f''(t)\,dt}\gt{0},$ so ${f'(1)\gt f'(0)=1},$ and $f'(1)$ cannot be $0.$

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