Fundamental Theorem of Calculus, Part 2

The fundamental theorem of calculus, part 2 says that if f is continuous on [a, b] and F is any antiderivative of f, then \int_a^b f(x)\,dx = F(b) - F(a). To evaluate a definite integral, find an antiderivative, plug in the top and bottom limits, and subtract. You don’t need a Riemann sum.

Before this lesson: Definite Integrals, Antiderivatives and Indefinite Integrals and Fundamental Theorem of Calculus, Part 1

What the theorem means

A definite integral adds up tiny pieces f(x) dx from a to b. If f is a rate of change, those pieces are tiny changes in an amount. Adding all the tiny changes gives the total change, which is the final amount minus the starting amount.

That is the whole idea. If F′ = f, then F is the amount and f is its rate, so \int_a^b f(x)\,dx = F(b) - F(a). The area under the curve equals the net change in any antiderivative.

Why it works: By Part 1, the area function G(x) = \int_a^x f(t)\,dt is an antiderivative of f. Any other antiderivative is F = G + C. So F(b) − F(a) = G(b) − G(a), and G(a) = 0, which leaves G(b), the integral from a to b.

Fundamental theorem of calculus part 2 formula

If f is continuous on [a, b] and F is any antiderivative of f, then:

\int_a^b f(x)\,dx = \Big[F(x)\Big]_a^b = F(b) - F(a)

The bracket with a and b is evaluation notation. It means “plug in b, plug in a, and subtract.”

How to evaluate a definite integral:

1. Check that f is continuous on [a, b].
2. Find an antiderivative F. You can leave off + C, because it cancels when you subtract.
3. Find F(b) and F(a).
4. Subtract: top value minus bottom value.

Fundamental theorem of calculus part 2 examples

Example 1: A power function

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

\int_1^3 x^2\,dx = \Big[\dfrac{x^3}{3}\Big]_1^3

= \dfrac{3^3}{3} - \dfrac{1^3}{3}

= \dfrac{27}{3} - \dfrac{1}{3}

= \dfrac{26}{3}

Answer: \dfrac{26}{3}, about 8.67. This is the shaded area in the graph above.

Example 2: A polynomial with a negative limit

Evaluate \int_{-1}^{2} (3x^2 - 4x + 5)\,dx.

F(x) = x^3 - 2x^2 + 5x

F(2) = 2^3 - 2(2)^2 + 5(2)

F(2) = 8 - 8 + 10 = 10

F(-1) = (-1)^3 - 2(-1)^2 + 5(-1)

F(-1) = -1 - 2 - 5 = -8

\int_{-1}^{2} (3x^2 - 4x + 5)\,dx = 10 - (-8)

= 18

Answer: 18. The parentheses around −1 and around −8 are what keep the signs right.

Example 3: A trig function

Evaluate \int_{\pi/6}^{\pi/2} \cos x\,dx.

\int_{\pi/6}^{\pi/2} \cos x\,dx = \Big[\sin x\Big]_{\pi/6}^{\pi/2}

= \sin\dfrac{\pi}{2} - \sin\dfrac{\pi}{6}

= 1 - \dfrac{1}{2}

= \dfrac{1}{2}

Answer: \dfrac{1}{2}. Keep the calculator in radian mode if you check this with one.

Example 4: A root and a reciprocal

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

\sqrt{x} + \dfrac{2}{x} = x^{1/2} + \dfrac{2}{x}

\int_1^4 \left(x^{1/2} + \dfrac{2}{x}\right)dx = \Big[\dfrac{2}{3}x^{3/2} + 2\ln x\Big]_1^4

= \left(\dfrac{2}{3}(4)^{3/2} + 2\ln 4\right) - \left(\dfrac{2}{3}(1)^{3/2} + 2\ln 1\right)

= \left(\dfrac{2}{3}(8) + 2\ln 4\right) - \left(\dfrac{2}{3} + 0\right)

= \dfrac{16}{3} - \dfrac{2}{3} + 2\ln 4

= \dfrac{14}{3} + 2\ln 4

\approx 7.44

Answer: \dfrac{14}{3} + 2\ln 4, about 7.44. The exact form is the better final answer unless a decimal is asked for.

Example 5: Drinks sold at a coffee shop

A coffee shop sells drinks at a rate of r(t) = 30 + 12t - 3t^2 drinks per hour, where t is the number of hours after 6 a.m. How many drinks does it sell from 6 a.m. to 10 a.m.?

\int_0^4 (30 + 12t - 3t^2)\,dt

= \Big[30t + 6t^2 - t^3\Big]_0^4

= \left(30(4) + 6(4)^2 - 4^3\right) - 0

= 120 + 96 - 64

= 152

Answer: about 152 drinks. The integral of a rate gives the total change, here the number of drinks sold.

Common mistakes

Related methods

The net change theorem

Since F′ is the rate of change of F, Part 2 can be written \int_a^b F'(x)\,dx = F(b) - F(a). The integral of a rate is the net change in the amount. If v(t) is a velocity, the integral of v from a to b is the change in position (see position, velocity and acceleration).

Splitting an integral first

For a piecewise function or an absolute value, split the interval where the formula changes. Then apply Part 2 to each piece and add (see properties of definite integrals).

When no antiderivative is easy to find

Some functions, like e^{-x^2}, have no antiderivative made of basic functions. Then you estimate the integral with Riemann sums or the trapezoidal rule.

Practice problems

1. Evaluate \int_0^3 (2x + 1)\,dx.
2. Evaluate \int_1^2 x^4\,dx.
3. Evaluate \int_{-2}^{1} (x^2 - 1)\,dx.
4. Evaluate \int_1^9 \dfrac{1}{\sqrt{x}}\,dx.
5. Evaluate \int_0^{\pi} \sin x\,dx.
6. Evaluate \int_0^{\pi/4} \sec^2 x\,dx.
7. Evaluate \int_0^1 e^x\,dx.
8. Evaluate \int_1^{e^2} \dfrac{3}{x}\,dx.
9. Evaluate \int_1^2 \dfrac{x^3 + 2}{x^2}\,dx.
10. A balloon fills with air at a rate of 6\sqrt{t} cubic inches per second, t seconds after you start. How much air goes in from t = 1 to t = 4?

\int_0^3 (2x + 1)\,dx = \Big[x^2 + x\Big]_0^3

= (3^2 + 3) - (0^2 + 0)

= 12 - 0 = 12

\int_1^2 x^4\,dx = \Big[\dfrac{x^5}{5}\Big]_1^2

= \dfrac{2^5}{5} - \dfrac{1^5}{5}

= \dfrac{32}{5} - \dfrac{1}{5} = \dfrac{31}{5}

\int_{-2}^{1} (x^2 - 1)\,dx = \Big[\dfrac{x^3}{3} - x\Big]_{-2}^{1}

= \left(\dfrac{1^3}{3} - 1\right) - \left(\dfrac{(-2)^3}{3} - (-2)\right)

= \left(\dfrac{1}{3} - 1\right) - \left(-\dfrac{8}{3} + 2\right)

= -\dfrac{2}{3} - \left(-\dfrac{2}{3}\right)

= 0

\dfrac{1}{\sqrt{x}} = x^{-1/2}

\int_1^9 x^{-1/2}\,dx = \Big[2x^{1/2}\Big]_1^9

= 2\sqrt{9} - 2\sqrt{1}

= 6 - 2 = 4

\int_0^{\pi} \sin x\,dx = \Big[-\cos x\Big]_0^{\pi}

= -\cos\pi - (-\cos 0)

= -(-1) - (-1)

= 1 + 1 = 2

\int_0^{\pi/4} \sec^2 x\,dx = \Big[\tan x\Big]_0^{\pi/4}

= \tan\dfrac{\pi}{4} - \tan 0

= 1 - 0 = 1

\int_0^1 e^x\,dx = \Big[e^x\Big]_0^1

= e^1 - e^0

= e - 1

\approx 1.718

\int_1^{e^2} \dfrac{3}{x}\,dx = \Big[3\ln x\Big]_1^{e^2}

= 3\ln e^2 - 3\ln 1

= 3(2) - 3(0)

= 6

\dfrac{x^3 + 2}{x^2} = x + 2x^{-2}

\int_1^2 (x + 2x^{-2})\,dx = \Big[\dfrac{x^2}{2} - \dfrac{2}{x}\Big]_1^2

= \left(\dfrac{4}{2} - \dfrac{2}{2}\right) - \left(\dfrac{1}{2} - 2\right)

= 1 - \left(-\dfrac{3}{2}\right)

= \dfrac{5}{2}

\int_1^4 6t^{1/2}\,dt = \Big[4t^{3/2}\Big]_1^4

= 4(4)^{3/2} - 4(1)^{3/2}

= 4(8) - 4(1)

= 32 - 4 = 28

Fundamental theorem of calculus part 2 FAQ

What does the fundamental theorem of calculus part 2 say?

If f is continuous on [a, b] and F′ = f, then the integral of f from a to b equals F(b) − F(a). You evaluate a definite integral with any antiderivative.

Why don’t you add + C to a definite integral?

The constant shows up in both F(b) and F(a), so it cancels when you subtract. Any antiderivative gives the same answer.

Can a definite integral be negative?

Yes. Area below the x-axis counts as negative. If more of the area is below the axis than above it, the integral is negative.

What is the net change theorem?

It is Part 2 applied to a rate: the integral of F′ from a to b equals F(b) − F(a). Integrating a rate of change gives the total change in the amount.

Summary

• Part 2: if F′ = f and f is continuous on [a, b], then \int_a^b f(x)\,dx = F(b) - F(a).
• Steps: find an antiderivative, plug in both limits, subtract top minus bottom.
• Leave off + C; it cancels.
• Put substituted values in parentheses, especially negative ones.
• The integral of a rate is the net change in the amount.
• The function must be continuous on the whole interval.

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