5 Integrals and Their Applications

A progressive guide to antiderivatives, definite integrals, Riemann sums, the Fundamental Theorem of Calculus, integral properties, and substitution techniques.

Antiderivatives and Indefinite Integrals

Integration answers two closely related questions: what function has a given rate of change, and how much total signed accumulation occurs over an interval? The first question leads to antiderivatives and indefinite integrals; the second leads to definite integrals.

An of $f$ is a function $F$ satisfying

F'(x)=f(x).

Because the derivative of a constant is zero, all antiderivatives of the same function differ by a constant. Thus, the indefinite integral is the family

\int f(x)\,dx=F(x)+C.

The expression being integrated is the integrand, $x$ is the variable of integration, and $C$ is the .

Basic rules

For $n\neq -1$ , the power rule is

\int x^n\,dx=\frac{x^{n+1}}{n+1}+C.

Other useful rules are

\int k\,dx=kx+C, \qquad \int \frac{1}{x}\,dx=\ln|x|+C,

\int e^x\,dx=e^x+C, \qquad \int \cos x\,dx=\sin x+C, \qquad \int \sin x\,dx=-\cos x+C.

Integration is linear, so constants and sums can be handled term by term:

\int [af(x)+bg(x)]\,dx =a\int f(x)\,dx+b\int g(x)\,dx.

This rule does not generally extend to products or quotients.

Example

\begin{aligned} \int(6x^2-4x+5)\,dx &=6\left(\frac{x^3}{3}\right)-4\left(\frac{x^2}{2}\right)+5x+C\\ &=2x^3-2x^2+5x+C. \end{aligned}

Differentiating the result gives $6x^2-4x+5$ , confirming the calculation.

Takeaway: An indefinite integral gives a family of antiderivatives, so the constant $C$ must be included.

Definite Integrals and Riemann Sums

A has limits of integration and produces a number:

\int_a^b f(x)\,dx.

The limits $a$ and $b$ specify the interval, and $dx$ indicates that accumulation is measured with respect to $x$ . A represents signed area or net change. Portions of the graph above the $x$ -axis contribute positively, and portions below the $x$ -axis contribute negatively.

If $f$ is nonnegative throughout $[a,b]$ , then the equals the geometric area between the graph and the $x$ -axis. If $f$ changes sign, positive and negative contributions cancel, so the result is net area rather than total geometric area.

From rectangles to an exact value

Partition $[a,b]$ into $n$ equal subintervals. Each has width

\Delta x=\frac{b-a}{n}.

Choose a sample point $x_i^*$ in each subinterval. The resulting is

\sum_{i=1}^{n}f(x_i^*)\Delta x.

As the number of rectangles increases and their widths approach zero, the approximation becomes the exact integral:

\int_a^b f(x)\,dx =\lim_{n\to\infty}\sum_{i=1}^{n}f(x_i^*)\Delta x.

Sample points may be chosen at left endpoints, right endpoints, or midpoints. For example, for $f(x)=x^2$ on $[0,2]$ , a right-endpoint sum uses

\Delta x=\frac{2}{n}, \qquad x_i^*=\frac{2i}{n},

so the approximation is

\sum_{i=1}^{n}\left(\frac{2i}{n}\right)^2\frac{2}{n}.

The exact value is

\int_0^2x^2\,dx =\left.\frac{x^3}{3}\right|_0^2 =\frac{8}{3}.

Takeaway: A is the limit of increasingly accurate signed-area approximations.

The

The gives two complementary ways to connect accumulation and rates of change.

Accumulation functions

For a continuous function $f$ , define an by

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

The variable $t$ is a dummy variable inside the integral. The endpoint $x$ controls how far the accumulation extends. The first part of the states

A'(x)=\frac{d}{dx}\int_a^x f(t)\,dt=f(x).

Thus, differentiating an recovers the original rate of accumulation.

If the upper limit is a function of $x$ , apply the chain rule:

\frac{d}{dx}\int_a^{g(x)}f(t)\,dt=f(g(x))g'(x).

For example, if

B(x)=\int_2^x\sqrt{1+t^3}\,dt,

then

B'(x)=\sqrt{1+x^3}.

An elementary formula for $B(x)$ is not required in order to find its derivative.

Evaluating definite integrals

The second part of the theorem states that if $F'(x)=f(x)$ , then

\int_a^b f(x)\,dx=F(b)-F(a).

This is often written as $[F(x)]_a^b$ . For example,

\begin{aligned} \int_1^3(2x+1)\,dx &=\left[x^2+x\right]_1^3\\ &=(9+3)-(1+1)\\ &=10. \end{aligned}

Takeaway: The first part differentiates accumulation functions; the second evaluates definite integrals using antiderivatives.

Properties of Definite Integrals

Definite integrals obey rules that make them easier to simplify, compare, and estimate.

Orientation and interval structure

A zero-length interval contributes nothing:

\int_a^a f(x)\,dx=0.

Reversing the limits changes the sign:

\int_b^a f(x)\,dx=-\int_a^b f(x)\,dx.

Adjacent intervals can be combined:

\int_a^b f(x)\,dx+ \int_b^c f(x)\,dx =\int_a^c f(x)\,dx.

More generally, for any point $c$ ,

\int_a^b f(x)\,dx =\int_a^c f(x)\,dx+ \int_c^b f(x)\,dx.

Linearity and comparison

Constants can be taken outside an integral, and sums can be separated:

\int_a^b[cf(x)+g(x)]\,dx =c\int_a^b f(x)\,dx+ \int_a^b g(x)\,dx.

If $f(x)\ge 0$ on $[a,b]$ , then

\int_a^b f(x)\,dx\ge 0.

If $f(x)\ge g(x)$ throughout the interval, then

\int_a^b f(x)\,dx\ge\int_a^b g(x)\,dx.

Bounds on a function produce bounds on its integral. If $m\le f(x)\le M$ , then

m(b-a)\le\int_a^b f(x)\,dx\le M(b-a).

These inequalities are useful for checking whether a computed answer is plausible.

Takeaway: The order of limits controls sign, intervals can be split or joined, and pointwise bounds lead to integral bounds.

as Reverse Chain Rule

reverses the chain rule. It is useful when an integrand contains a composite expression together with its derivative, or a close multiple of that derivative.

Set

u=g(x), \qquad du=g'(x)\,dx.

Then an integral of the form

\int f(g(x))g'(x)\,dx

becomes

\int f(u)\,du.

For an indefinite integral, integrate in terms of $u$ , then substitute back.

Indefinite example

Evaluate

\int 2x(x^2+4)^5\,dx.

Choose

u=x^2+4, \qquad du=2x\,dx.

The integral becomes

\begin{aligned} \int 2x(x^2+4)^5\,dx &=\int u^5\,du\\ &=\frac{u^6}{6}+C\\ &=\frac{(x^2+4)^6}{6}+C. \end{aligned}

Definite

For a , change the limits along with the variable:

\int_a^b f(g(x))g'(x)\,dx =\int_{g(a)}^{g(b)}f(u)\,du.

Consider

\int_0^1 2x\sqrt{x^2+1}\,dx.

Let $u=x^2+1$ , so $du=2x\,dx$ . The limits become $u=1$ when $x=0$ and $u=2$ when $x=1$ . Therefore,

\begin{aligned} \int_0^1 2x\sqrt{x^2+1}\,dx &=\int_1^2u^{1/2}\,du\\ &=\left[\frac{2}{3}u^{3/2}\right]_1^2\\ &=\frac{2}{3}(2^{3/2}-1). \end{aligned}

Recognizing the pattern

Look for a complicated inner expression whose derivative also appears elsewhere in the integrand. Common signals include a power of an expression multiplied by its derivative, an exponential multiplied by the derivative of its exponent, or a denominator whose derivative appears in the numerator.

For example,

\int\frac{3x^2}{x^3+7}\,dx

suggests $u=x^3+7$ , because $du=3x^2\,dx$ . Thus,

\int\frac{3x^2}{x^3+7}\,dx =\int\frac{1}{u}\,du =\ln|u|+C =\ln|x^3+7|+C.

Takeaway: Choose the inner expression as $u$ , account for its derivative, and change definite-integral limits immediately.

Integrated Problem-Solving Summary

The main ideas form one connected workflow:

Find an when the goal is to recover a function from its derivative.
Use a to represent signed accumulation over an interval.
Understand a as the limit of Riemann sums.
Use the to differentiate accumulation functions or evaluate definite integrals.
Apply the properties of definite integrals to split intervals, reverse limits, compare values, and estimate results.
Use when a composite expression and its derivative appear together.

The central formulas are

\int f(x)\,dx=F(x)+C, \qquad F'(x)=f(x),

\frac{d}{dx}\int_a^x f(t)\,dt=f(x), \qquad \int_a^b f(x)\,dx=F(b)-F(a),

and

\int_a^b f(g(x))g'(x)\,dx =\int_{g(a)}^{g(b)}f(u)\,du.

Always distinguish an indefinite integral, which is a family of functions, from a , which is a number. For definite integrals, keep track of sign, interval orientation, and any changes to limits made during .