Fundamental Theorem of Calculus, Part 1

The fundamental theorem of calculus, part 1 says that if f is continuous, the accumulation function F(x) = \int_a^x f(t)\,dt has derivative F'(x) = f(x). In words, the rate at which area builds up under a curve equals the height of the curve. Taking the derivative of an integral gives back the function inside.

Before this lesson: Definite Integrals, Properties of Definite Integrals, Antiderivatives and Indefinite Integrals and Chain Rule

What the theorem means

Start at a fixed point a and let the right edge x move. The signed area under f from a to x is a new function of x, called an accumulation function. We write it F(x) = \int_a^x f(t)\,dt. The letter t is a dummy variable: it only names the input inside the integral, so x can name the moving edge.

Now push the edge from x to x + h. The new area is a thin strip about f(x) tall and h wide, so F(x + h) - F(x) \approx f(x)\cdot h. Divide by h and let h shrink to 0. The difference quotient becomes F′(x), and the right side becomes f(x).

So the area function grows fastest where f is tallest. It shrinks where f is below the axis, because that area counts as negative.

Fundamental theorem of calculus part 1 formula

If f is continuous on an interval that contains a, then for every x in that interval:

\dfrac{d}{dx}\int_a^x f(t)\,dt = f(x)

When the upper limit is a function g(x), use the chain rule:

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

When both limits change, subtract the lower limit’s term:

\dfrac{d}{dx}\int_{h(x)}^{g(x)} f(t)\,dt = f\big(g(x)\big)\,g'(x) - f\big(h(x)\big)\,h'(x)

How to use it:

1. Check that the constant limit is a number and that f is continuous.
2. If x is the lower limit, flip the limits and add a minus sign.
3. Replace t with the upper limit in f.
4. Multiply by the derivative of the upper limit.

Fundamental theorem of calculus part 1 examples

Example 1: Upper limit x

Let F(x) = \int_2^x (t^3 + 5t)\,dt. Find F′(x) and F(2).

F'(x) = \dfrac{d}{dx}\int_2^x (t^3 + 5t)\,dt

F'(x) = x^3 + 5x

F(2) = \int_2^2 (t^3 + 5t)\,dt

F(2) = 0

Answer: F'(x) = x^3 + 5x and F(2) = 0. The lower limit 2 does not change the derivative. It only sets where F equals 0.

Example 2: Upper limit x²

Find \dfrac{d}{dx}\int_1^{x^2} \sqrt{1 + t^3}\,dt.

f(t) = \sqrt{1 + t^3},\; g(x) = x^2

f\big(g(x)\big) = \sqrt{1 + (x^2)^3}

f\big(g(x)\big) = \sqrt{1 + x^6}

g'(x) = 2x

\dfrac{d}{dx}\int_1^{x^2} \sqrt{1 + t^3}\,dt = \sqrt{1 + x^6}\cdot 2x

= 2x\sqrt{1 + x^6}

Answer: 2x\sqrt{1 + x^6}. We never had to find an antiderivative of \sqrt{1 + t^3}, which has no simple formula.

Example 3: x in the lower limit

Find \dfrac{d}{dx}\int_x^4 \cos(t^2)\,dt.

\int_x^4 \cos(t^2)\,dt = -\int_4^x \cos(t^2)\,dt

\dfrac{d}{dx}\left(-\int_4^x \cos(t^2)\,dt\right) = -\cos(x^2)

Answer: -\cos(x^2). When x is the lower limit, the answer has a minus sign.

Example 4: Both limits change

Find \dfrac{d}{dx}\int_{2x}^{x^3} \dfrac{1}{1 + t^2}\,dt.

\int_{2x}^{x^3} \dfrac{dt}{1 + t^2} = \int_0^{x^3} \dfrac{dt}{1 + t^2} - \int_0^{2x} \dfrac{dt}{1 + t^2}

\dfrac{d}{dx}\int_0^{x^3} \dfrac{dt}{1 + t^2} = \dfrac{1}{1 + x^6}\cdot 3x^2

\dfrac{d}{dx}\int_0^{2x} \dfrac{dt}{1 + t^2} = \dfrac{1}{1 + 4x^2}\cdot 2

\dfrac{3x^2}{1 + x^6} - \dfrac{2}{1 + 4x^2}

Answer: \dfrac{3x^2}{1 + x^6} - \dfrac{2}{1 + 4x^2}. This matches the both-limits formula: top term minus bottom term.

Example 5: Water in a tank

A tank holds 50 liters at time 0. Water flows at a rate of 6 − 2t liters per minute, t minutes later. Negative values mean water drains out. The volume after x minutes is V(x) = 50 + \int_0^x (6 - 2t)\,dt. Find the rate of change at 1 minute and at 5 minutes. Then find when the tank is fullest and how much it holds then.

V'(x) = 0 + (6 - 2x)

V'(1) = 6 - 2(1) = 4

V'(5) = 6 - 2(5) = -4

6 - 2x = 0

2x = 6

x = 3

V'(x) \gt 0 \text{ for } x \lt 3,\; V'(x) \lt 0 \text{ for } x \gt 3

\int_0^3 (6 - 2t)\,dt = \tfrac{1}{2}(3)(6)

\tfrac{1}{2}(3)(6) = 9

V(3) = 50 + 9 = 59

Answer: the volume rises at 4 liters per minute at 1 minute and falls at 4 liters per minute at 5 minutes. The tank is fullest at 3 minutes, with 59 liters.

Common mistakes

Related methods

Reading an accumulation function from a graph

If g(x) = \int_a^x f(t)\,dt, then g′ = f and g″ = f′. So g increases where the graph of f is above the axis and decreases where it is below. g has a local maximum where f crosses from positive to negative. g has an inflection point where f has a peak or a valley (see concavity and inflection points).

Position from velocity

If v(t) is a velocity, then s(x) = s(0) + \int_0^x v(t)\,dt is the position, and Part 1 gives s′(x) = v(x). This links integrals to position, velocity and acceleration.

Part 2 of the theorem

Part 1 shows that \int_a^x f(t)\,dt is an antiderivative of f. Part 2 turns that fact into a shortcut: evaluate an antiderivative at the two limits and subtract. That is the next lesson, Fundamental Theorem of Calculus, Part 2.

Practice problems

1. Find \dfrac{d}{dx}\int_1^x (t^2 - 3t)\,dt.
2. Find \dfrac{d}{dx}\int_0^x \sqrt{t + 4}\,dt.
3. Find \dfrac{d}{dx}\int_5^x e^{-t^2}\,dt.
4. Find \dfrac{d}{dx}\int_x^2 t\sin t\,dt.
5. Find \dfrac{d}{dx}\int_0^{x^3} \cos t\,dt.
6. Find \dfrac{d}{dx}\int_1^{\sqrt{x}} t^2\,dt for positive x.
7. Find \dfrac{d}{dx}\int_{x^2}^{0} \dfrac{1}{1 + t}\,dt.
8. Find \dfrac{d}{dx}\int_x^{3x} t^2\,dt.
9. Let g(x) = \int_0^x (t - 2)\,dt. Find g′(x) and g″(x). Where does g decrease, and what is its smallest value?
10. Snow falls on a roof at a rate of 0.5 + 0.25t inches per hour, t hours after it starts. The depth after x hours is S(x) = \int_0^x (0.5 + 0.25t)\,dt. Find S′(4) and S(4).

\dfrac{d}{dx}\int_1^x (t^2 - 3t)\,dt = x^2 - 3x

\dfrac{d}{dx}\int_0^x \sqrt{t + 4}\,dt = \sqrt{x + 4}

\dfrac{d}{dx}\int_5^x e^{-t^2}\,dt = e^{-x^2}

\int_x^2 t\sin t\,dt = -\int_2^x t\sin t\,dt

\dfrac{d}{dx}\left(-\int_2^x t\sin t\,dt\right) = -x\sin x

\cos(x^3)

\dfrac{d}{dx}(x^3) = 3x^2

\cos(x^3)\cdot 3x^2 = 3x^2\cos(x^3)

(\sqrt{x})^2 = x

\dfrac{d}{dx}\sqrt{x} = \dfrac{1}{2\sqrt{x}}

x\cdot\dfrac{1}{2\sqrt{x}} = \dfrac{x}{2\sqrt{x}}

\dfrac{x}{2\sqrt{x}} = \dfrac{\sqrt{x}}{2}

\int_{x^2}^{0} \dfrac{dt}{1 + t} = -\int_0^{x^2} \dfrac{dt}{1 + t}

\dfrac{1}{1 + x^2}\cdot 2x

-\dfrac{2x}{1 + x^2}

(3x)^2\cdot 3 = 9x^2\cdot 3 = 27x^2

x^2\cdot 1 = x^2

27x^2 - x^2 = 26x^2

g'(x) = x - 2

g''(x) = 1

x - 2 \lt 0 \text{ when } x \lt 2

x - 2 \gt 0 \text{ when } x \gt 2

g(2) = \int_0^2 (t - 2)\,dt = -\tfrac{1}{2}(2)(2)

g(2) = -2

S'(x) = 0.5 + 0.25x

S'(4) = 0.5 + 0.25(4) = 0.5 + 1 = 1.5

S(4) = \tfrac{1}{2}(0.5 + 1.5)(4)

S(4) = \tfrac{1}{2}(2)(4) = 4

Fundamental theorem of calculus part 1 FAQ

What does the fundamental theorem of calculus part 1 say?

If f is continuous, then \dfrac{d}{dx}\int_a^x f(t)\,dt = f(x). Taking the derivative of an accumulation function gives back the function being integrated.

What is the difference between part 1 and part 2?

Part 1 is about taking the derivative of an integral with a variable limit. Part 2 is about evaluating a definite integral: find an antiderivative F, then compute F(b) − F(a).

How do you use the chain rule with the fundamental theorem of calculus?

If the upper limit is g(x), replace t with g(x) and multiply by g′(x). For example, \dfrac{d}{dx}\int_0^{x^2} \sin t\,dt = 2x\sin(x^2).

Why is the variable inside the integral t instead of x?

The letter x already names the moving limit. Using t inside keeps the two roles apart, and t disappears once the integral is computed.

Summary

• Part 1: \dfrac{d}{dx}\int_a^x f(t)\,dt = f(x) when f is continuous.
• An accumulation function’s rate of change is the height of the curve at its right edge.
• Upper limit g(x): multiply f(g(x)) by g′(x).
• x in the lower limit: flip the limits and add a minus sign.
• Both limits change: top term minus bottom term.
• The constant limit only sets where the function equals 0.

Previous lesson: Properties of Definite Integrals

Next lesson: Fundamental Theorem of Calculus, Part 2

Back to the Calculus 1 Index