The area function and the Fundamental Theorem of Calculus

What you will be able to do

• say what a definite integral measures, and why it can be negative;
• explain why the derivative of the area function is the function itself;
• evaluate a definite integral with an antiderivative.

Distance from speed

A bus moves on a straight road. At a steady speed of $12\ \text{m/s}$, it covers $12\ \text{m}$ each second, so in $5\ \text{s}$ it covers $60\ \text{m}$. On the speed graph this is the area of a rectangle, $12 \times 5$. When the speed changes, the distance is still the area under the speed graph, but the shape is no longer a rectangle.

The area under a curve $y = f(x)$ from $x = a$ to $x = b$ is written

The numbers $a$ and $b$ are the limits of the integral. The letter $x$ is a dummy variable: $\int_a^b f(x)\,dx$ and $\int_a^b f(t)\,dt$ are the same number. Where $f$ is below the axis, the area counts as negative. So the definite integral is a signed area. The area of a region is its size, always positive, and it needs care when $f$ changes sign.

The area function

Fix a start point $a$ and let the end point move. The area from $a$ to $x$ is a function of $x$:

The simulation shows it for a bus with speed $f(t) = 4t - t^2$. The orange curve is the distance travelled, and the blue curve is the speed.

The Fundamental Theorem of Calculus

You saw in the simulation that the slope of the orange curve at $b$ equals the height of the blue curve at $b$. This is no accident.

Part 1. If $f$ is continuous on an interval and $A(x) = \int_a^x f(t)\,dt$, then $A'(x) = f(x)$.

Here is why. When $x$ grows by a small amount $h$, the area grows by a thin strip of width $h$ and height about $f(x)$. So $A(x + h) - A(x) \approx f(x)\,h$. Divide by $h$ and let $h \to 0$. The slope of $A$ is $f(x)$. So $A$ is an antiderivative of $f$.

Part 2. If $F$ is any antiderivative of $f$, then

Here is why. $A$ and $F$ are both antiderivatives of $f$, so $A(x) = F(x) + C$. Since $A(a) = 0$, we get $C = -F(a)$. Then $A(b) = F(b) - F(a)$.

We write $F(b) - F(a)$ as $\left[F(x)\right]_a^b$. The constant $C$ cancels, so we do not need it in a definite integral.

Worked examples

Example 1. Find $\displaystyle\int_1^3 (2x + 1)\,dx$.

1. An antiderivative is $F(x) = x^2 + x$.
2. $F(3) - F(1) = (9 + 3) - (1 + 1) = 10$.

Example 2. Find $\displaystyle\int_0^{\pi/2} \cos x\,dx$ and $\displaystyle\int_0^1 e^x\,dx$.

1. $\left[\sin x\right]_0^{\pi/2} = 1 - 0 = 1$.
2. $\left[e^x\right]_0^1 = e - 1 = 1.7183$.

Example 3 (signed area). Find $\displaystyle\int_0^{2\pi} \sin x\,dx$, and the area between the sine curve and the $x$-axis from $0$ to $2\pi$.

1. $\left[-\cos x\right]_0^{2\pi} = -1 - (-1) = 0$. The positive hump and the negative hump cancel.
2. The area is the size of each hump added: $2 \times \displaystyle\int_0^{\pi}\sin x\,dx = 2 \times 2 = 4$.

When the theorem does not apply

The theorem needs $f$ to be continuous on $[a, b]$. Take $\displaystyle\int_{-1}^{1}\dfrac{dx}{x^2}$. The antiderivative is $-\dfrac{1}{x}$, and putting in the limits gives $-1 - 1 = -2$. This is wrong. The integrand is never negative, so no area can be negative. The function $\dfrac{1}{x^2}$ is not defined at $x = 0$, and it grows without bound near $0$. The theorem cannot be used across the break.

Practise with new numbers

Try it in Python

The program finds each area by adding many thin strips, the midpoint rule, and compares it with $F(b) - F(a)$.

Common mistakes

• Forgetting $F(a)$. The answer is $F(b) - F(a)$, not $F(b)$.
• Subtracting the wrong way round. Upper limit first, then take away the lower limit.
• Treating signed area as area. If $f$ changes sign, split the interval where $f = 0$ and add the sizes.
• Using the theorem across a break. The integrand must be continuous on the whole interval.
• Adding $C$ and then keeping it. In a definite integral the constant cancels.

Remember these

• $\int_a^b f(x)\,dx$ is the signed area. It is negative where $f$ is below the axis.
• Part 1: if $A(x) = \int_a^x f(t)\,dt$, then $A'(x) = f(x)$.
• Part 2: $\int_a^b f(x)\,dx = F(b) - F(a)$.
• The theorem needs $f$ continuous on $[a, b]$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find $\displaystyle\int_1^4 \sqrt{x}\,dx$.

2. Find $\displaystyle\int_0^{\pi}\sin x\,dx$.

3. Find $\displaystyle\int_0^1 \dfrac{dx}{1 + x^2}$.

4. Let $A(x) = \displaystyle\int_0^x t^3\,dt$. What is $A'(x)$, and what is $A'(2)$?

5. A student writes $\displaystyle\int_{-2}^{2}\dfrac{dx}{x^2} = \left[-\dfrac{1}{x}\right]_{-2}^{2} = -1$. What is wrong?

References

• OpenStax, Calculus Volume 1, Chapter 5, section 5.3 The Fundamental Theorem of Calculus, and section 5.2 The Definite Integral.
• NCERT, Mathematics Part II for Class XII, Chapter 7, the sections on the definite integral and the Fundamental Theorem of Calculus.