The fundamental theorem and the power rule for a parabola, integrals from easy to advanced lesson

The region under $y = x^2$ from $0$ to $2$ has a curved top, so no rectangle or triangle formula gives its area, and a limit of Riemann sums would need a formula for a sum of squares. The question is the value of

\int_0^2 x^2\,dx.

This page lets the right endpoint move, shows that the accumulated area grows at a rate equal to the height of the curve, and uses the fundamental theorem of calculus with the antiderivative $\tfrac13 x^3$ to get exactly $\tfrac83$ . It also separates that number from the indefinite integral, a family of functions that carries a constant of integration.

Key terms

Accumulation function

Definition Let $f$ be continuous on an interval $I$ and let $a$ be a point of $I$ . The accumulation function of $f$ from $a$ is the function

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

defined for every $x$ in $I$ , where for $x < a$ the integral means $-\int_x^a f(t)\,dt$ . It starts at $A(a) = 0$ . The moving endpoint $x$ is the input of $A$ , so the variable of integration gets a different letter.

In plain words Move the edge of the region from $a$ to $x$ and record the signed area swept out. The integrand gives the height at the moving edge, and the accumulation function gives the total collected so far. They are different functions, and the fundamental theorem below says how they are related.

Example For the constant $f(t) = 2$ and $a = 0$ , the region up to a point $x \ge 0$ is a rectangle of width $x$ and height $2$ , so $A(x) = 2x$ . For $f(t) = t$ it is a triangle with base $x$ and height $x$ , so $A(x) = \tfrac12 x^2$ . In both cases $A'(x) = f(x)$ , the height at the moving edge. For $f(t) = t^2$ the accumulation function

A(x) = \int_0^x t^2\,dt

starts at $A(0) = 0$ , and its value $A(2)$ is the area under the parabola that this page is after.

Antiderivative

Definition Let $f$ be defined on an open interval $I$ . A function $F$ is an antiderivative of $f$ on $I$ if $F'(x) = f(x)$ for every $x$ in $I$ . If $F$ is one antiderivative of $f$ on $I$ , then the antiderivatives of $f$ on $I$ are exactly the functions $F(x) + C$ with $C$ a constant. The reason is the mean value theorem, by which a function whose derivative is $0$ at every point of an interval is constant on that interval.

In plain words An antiderivative undoes differentiation. Finding one is a guess that can always be checked, since differentiating the candidate must give back the integrand exactly.

Example Differentiating $x^3$ gives $3x^2$ , three times too much, so divide by $3$ . The function $F(x) = \tfrac13 x^3$ has $F'(x) = \tfrac13 \cdot 3x^2 = x^2$ , so it is an antiderivative of $x^2$ on the whole real line. So are $\tfrac13 x^3 + 5$ and $\tfrac13 (x^3 + 6) = \tfrac13 x^3 + 2$ . The function $x^3$ itself is not, because its derivative is $3x^2$ .

Fundamental theorem of calculus

Definition Let $f$ be continuous on $[a, b]$ with $a < b$ . The first part of the theorem says that the accumulation function $A(x) = \int_a^x f(t)\,dt$ is continuous on $[a, b]$ and differentiable on $(a, b)$ , with

A'(x) = f(x).

The second part says that if $F$ is continuous on $[a, b]$ and $F'(x) = f(x)$ for every $x$ in $(a, b)$ , for example an antiderivative of $f$ on an open interval containing $[a, b]$ , then

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

The difference $F(b) - F(a)$ is also written $\bigl[F(x)\bigr]_a^b$ .

In plain words The accumulated area grows at a rate equal to the height of the graph at the moving edge. So when the integrand is continuous on $[a, b]$ , every function whose derivative is the integrand there changes by exactly the value of the integral between the two limits, and two evaluations of that function replace the limit of sums.

Example For $\int_0^2 x^2\,dx$ , the integrand $f(x) = x^2$ is a polynomial, so it is continuous on $[0, 2]$ . Let $A(x) = \int_0^x t^2\,dt$ . For $x \ge 0$ and $h > 0$ the strip added between $x$ and $x + h$ lies between the rectangles of heights $x^2$ and $(x + h)^2$ on that piece, so

\begin{aligned} x^2 &\le \frac{A(x + h) - A(x)}{h} \\[0.4em] &\le (x + h)^2. \end{aligned}

Letting $h$ shrink to $0$ squeezes the average height to $x^2$ , and a small negative $h$ works the same way, which gives $A'(x) = x^2$ . The second part then applies with $F(x) = \tfrac13 x^3$ , whose derivative is $x^2$ on the whole real line,

\begin{aligned} \int_0^2 x^2\,dx &= \left[\frac{x^3}{3}\right]_0^2 \\[0.4em] &= \frac83 - 0 = \frac83. \end{aligned}

The value $\tfrac83 \approx 2.67$ is the area, while $f(2) = 4$ is only the height of the curve at the right edge. As a check on size, $2x - x^2 = x(2 - x) > 0$ for $0 < x < 2$ , so the parabola lies below the line $y = 2x$ there, and the area lies between $0$ and the area $4$ of the triangle under that line. The answer is a number and carries no $+ C$ .

Indefinite integral and the constant of integration

Definition Let $F$ be an antiderivative of $f$ on an open interval $I$ . The indefinite integral of $f$ is the family of all antiderivatives of $f$ on $I$ , written

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

where the constant of integration $C$ ranges over all real numbers. An indefinite integral is a family of functions, while a definite integral $\int_a^b f(x)\,dx$ is one number. On a domain made of separate intervals, such as all $x \ne 0$ , the constant may take a different value on each interval.

In plain words The members of the family differ by vertical shifts. Shifting a graph up or down leaves its slope at every $x$ unchanged, so every member has the same derivative. The $+ C$ records that differentiation forgets constants.

Example The antiderivative found above gives

\int x^2\,dx = \frac{x^3}{3} + C.

Every member gives the same definite integral over $[0, 2]$ , because the constant cancels in the subtraction,

\left(\frac83 + C\right) - (0 + C) = \frac83,

so the definite answer carries no $+ C$ . A condition picks out one member. The antiderivative of $x^2$ with value $1$ at $x = 0$ is $\tfrac13 x^3 + 1$ . On the domain $x \ne 0$ , the function equal to $-\tfrac1x + 1$ for $x > 0$ and to $-\tfrac1x - 4$ for $x < 0$ has derivative $x^{-2}$ everywhere on that domain, though it is not $-\tfrac1x + C$ for a single constant $C$ .

Reverse power rule

Definition Let $n$ be a real number other than $-1$ . Then

\frac{d}{dx}\,\frac{x^{n+1}}{n+1} = x^n,

so $x^{n+1}/(n+1)$ is an antiderivative of $x^n$ and

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

This holds for all real $x$ when $n$ is a nonnegative integer, on every open interval that avoids $0$ when $n$ is a negative integer, and for $x > 0$ when $n$ is not an integer. The exponent $n = -1$ is excluded because the formula would divide by $0$ . Instead, $\tfrac1x$ has the antiderivative $\ln\lvert x \rvert$ on each interval that avoids $0$ .

In plain words To integrate a power of $x$ , raise the exponent by one and divide by the new exponent. Differentiating brings the new exponent down as a factor and lowers the power by one, which undoes both steps.

Example Raising the exponent of $x^5$ by one gives $x^6$ , and dividing by $6$ gives

\int x^5\,dx = \frac{x^6}{6} + C.

The rule also covers roots. The function $\sqrt{x} = x^{1/2}$ is continuous on $[1, 4]$ and has the antiderivative $\tfrac23 x^{3/2}$ for $x > 0$ , so

\begin{aligned} \int_1^4 \sqrt{x}\,dx &= \frac23\bigl(4^{3/2} - 1^{3/2}\bigr) \\[0.4em] &= \frac23\,(8 - 1) = \frac{14}{3}. \end{aligned}

Check yourself

Evaluate $\int_1^3 x^2\,dx$ .

Show the answer

The integrand $x^2$ is continuous on $[1, 3]$ and $F(x) = \tfrac13 x^3$ is an antiderivative, so

\begin{aligned} \int_1^3 x^2\,dx &= \frac{27}{3} - \frac13 \\[0.4em] &= \frac{26}{3}. \end{aligned}

The lower value $F(1) = \tfrac13$ is not $0$ here, and leaving it out would give the wrong answer $9$ . As a check, the height runs from $1$ to $9$ over a width of $2$ , and $\tfrac{26}{3} \approx 8.67$ lies between $1 \cdot 2 = 2$ and $9 \cdot 2 = 18$ .

Let $A(x) = \int_1^x t^3\,dt$ . Find a formula for $A(x)$ , check that $A(1) = 0$ and $A'(x) = x^3$ , and say which member of the family $\int x^3\,dx$ it is.

Show the answer

By the reverse power rule, $\tfrac14 t^4$ is an antiderivative of $t^3$ on the whole real line, and $t^3$ is continuous everywhere, so the fundamental theorem gives

A(x) = \frac{x^4}{4} - \frac14

for every real $x$ . Then $A(1) = \tfrac14 - \tfrac14 = 0$ and $A'(x) = x^3$ , as the first part of the theorem says. It is the member of $\tfrac14 x^4 + C$ with $C = -\tfrac14$ , the one that vanishes at the base point $1$ .

A classmate writes $\int_{-1}^{1} x^{-2}\,dx = \bigl[-\tfrac1x\bigr]_{-1}^{1} = -1 - 1 = -2$ . Explain why this cannot be right and what happens to the integral.

Show the answer

The integrand $x^{-2}$ is positive wherever it is defined, so a negative answer is impossible. The fundamental theorem needs a function continuous on all of $[-1, 1]$ , but $x^{-2}$ is undefined at $0$ and grows without bound near it, and $-\tfrac1x$ is not an antiderivative on any interval containing $0$ . The area near $0$ is in fact infinite. For $0 < \varepsilon < 1$ ,

\int_\varepsilon^1 x^{-2}\,dx = \frac{1}{\varepsilon} - 1,

which grows without bound as $\varepsilon \to 0^{+}$ . So the integral has no finite value, and the same happens on the left of $0$ .

The fundamental theorem and the power rule for a parabola

Key terms

Accumulation function

Antiderivative

Fundamental theorem of calculus

Indefinite integral and the constant of integration

Reverse power rule

Check yourself

How this lesson connects

Builds on

Used later in