Evaluate at the bounds in the correct order

The area is the rise of F

The top graph shows \(f(x) = 2x + 1\). The area of the teal region, from \(x = 1\) to the orange edge, is the definite integral of \(f\) from 1 to the edge.

The lower graph shows \(\cb{F(x) = x^2 + x}\), an antiderivative of \(f\), so its slope at each \(x\) equals the height of the top curve.

The orange bracket shows how far \(F\) climbs from \(x = 1\) to the edge. \(F(3) = 12\) and \(F(1) = 2\), so the climb is \(\cy{12 - 2 = 10}\).

As a check, the region at (x = 3) is a trapezoid with parallel sides 3 and 7 and width 2, so its area is 10.

The green strip at the edge has area height times width. That height is also the slope of F, so F climbs by the same amount over the strip. When you drag the edge, the area and the climb grow together.

From the lower bound to the upper bound

The top graph shows a curve \(f\) and the region between it and the axis from \(a\) to \(b\). The orange bracket on the lower graph reaches from the height \(\cy{F(a)}\) of an antiderivative \(\cb{F}\) up to the height \(\cy{F(b)}\).

The bracket is as long as the area, for every a and b. Where the curve is below the axis, the region is red and counts as negative, so F falls and the bracket shortens. A shifted copy of F climbs by the same amount, so any antiderivative works.

Press Swap the bounds to start at \(b\) and finish at \(a\). A red bracket appears to the left of \(a\) and points down from \(F(b)\) to \(F(a)\), so it measures \(F(a) - F(b)\), the negative of the area.

The climb runs from the lower bound to the upper bound, so the value at the upper bound comes first.

Notation

1. Find an antiderivative \(\cb{F}\) of \(f\). The region starts at the lower bound \(a\) and ends at the upper bound \(\cy{b}\), the orange edge.

   \[\cb{F}'(x) = \cy{f(x)}\]

2. Substitute both bounds into \(F\). The upper value minus the lower value is the length of the orange bracket, which is the integral.

   \[\int_a^b f(x)\,dx = \cy{F(b) - F(a)}\]

3. Square brackets with the bounds record the two substitutions.

   \[\Big[\cb{F(x)}\Big]_a^b = \cy{F(b) - F(a)}\]

4. The other order gives the red bracket, which points down, so it is the negative of the integral.

   \[\cr{F(a) - F(b)} = -\int_a^b f(x)\,dx\]

5. Write a negative value in parentheses before you subtract it, because subtracting a negative number adds.

   \[5 - (-3) = 8\]

6. The function \(f\) must be continuous on the whole interval from \(a\) to \(b\), so its graph has no break there.

   The function \(1/x^2\) is not defined at 0. From \(-1\) to 2, the formula \(-\tfrac1x\) gives \(-\tfrac32\), which cannot be an area because the curve lies above the axis.

When it fails

The rule needs \(\cb{F}\) to have slope \(f(x)\) at every point of \([a, b]\). Put a jump in \(f\) at \(c\). Then move the edge \(\cy{b}\) past \(c\).

To the right of \(c\), the area grows faster. The red dashed curve is the old \(F\), which ignores the jump, so its rise is too small. At \(c\), no function has slope \(f(x)\), because a slope cannot jump.

Example 1: two substitutions

Evaluate \(\displaystyle\int_1^3 (x^2 + 2)\,dx\).

1. An antiderivative of \(x^2 + 2\) is \(\cb{F(x) = \tfrac{x^3}{3} + 2x}\), because its derivative is \(x^2 + 2\).

2. Substitute the upper bound 3.

   \[F(3) = 9 + 6 = 15\]

3. Substitute the lower bound 1.

   \[F(1) = \tfrac13 + 2 = \tfrac73\]

4. Subtract the lower value from the upper value.

   \[15 - \tfrac73 = \cy{\tfrac{38}{3}}\]

5. As a check, the region is 2 wide and 3 to 11 tall, so its area, about 12.7, lies between 6 and 22.

Example 2: a negative lower value

Evaluate \(\displaystyle\int_{-2}^{1} (4 - x^2)\,dx\).

1. An antiderivative of \(4 - x^2\) is \(\cb{F(x) = 4x - \tfrac{x^3}{3}}\).

2. Substitute the upper bound 1.

   \[F(1) = 4 - \tfrac13 = \tfrac{11}{3}\]

3. Substitute the lower bound \(-2\).

   \[F(-2) = -8 + \tfrac83 = -\tfrac{16}{3}\]

4. A tempting wrong move: Subtract the upper value from the lower value.

   \[\begin{aligned} F(-2) - F(1) &= -\tfrac{16}{3} - \tfrac{11}{3} \\ &= -9 \end{aligned}\]

   The region lies above the axis, but the result \(-9\) is negative. The red bracket points down, because this order measures the climb backward, from \(x = 1\) to \(x = -2\).

5. Subtract the lower value from the upper value, and put \(F(-2)\) in parentheses. The region is 3 wide and at most 4 tall, so 9 fits.

   \[\begin{aligned} F(1) - F(-2) &= \tfrac{11}{3} - \left(-\tfrac{16}{3}\right) \\ &= \cy{9} \end{aligned}\]

Practice

Evaluate \(\displaystyle\int_{-1}^{2} (3x^2 + 1)\,dx\).

Move the edge to 2. Substitute the upper bound first, and write \(F(-1)\) in parentheses. The region lies above the axis, so the answer is positive.

Questions

The top graph shows \(f(x) = x\). The lower graph shows the antiderivative \(F(x) = \tfrac12 x^2 + 1\).

1. Read \(F(0)\) and \(F(3)\) from the lower graph. What is \(\displaystyle\int_0^3 f(x)\,dx\)?

   • 4.5
   • 5.5
   • 6.5
   • \(-4.5\)
   Answer

   4.5

2. The lower bound is 4 and the upper bound is 0. What is \(\displaystyle\int_4^0 f(x)\,dx\)?

   • \(-8\)
   • 8
   • 9
   • \(-9\)
   Answer

   \(-8\)

3. What is \(\displaystyle\int_{-1}^{2} 5x^4\,dx\)?

   • 33
   • 31
   • 32
   • \(-33\)
   Answer

   33

Summary

You find an antiderivative, substitute the upper bound and then the lower bound, and subtract the lower value from the upper value.

It works because an antiderivative climbs by the area of the region, and the climb is the upper height minus the lower height.

You use it for any definite integral of a continuous function with a known antiderivative.

This is one of four skills in topic 6.7 of Which & Why: AP Calculus.