The Fundamental Theorem of Calculus

Slope and area, undone

Derivatives are slopes. Integrals are areas. And the Fundamental Theorem of Calculus says that differentiating and integrating undo each other. So slope and area are opposite operations, in the way that adding and subtracting are.

That should seem strange. A slope describes steepness at a single point. An area is a total over a whole interval. Why would one of them undo the other? This lesson is about why that is true, and why it earns the word fundamental.

Drag the ochre edge to sweep out area under the top graph, starting from a. The bottom graph records the area so far, A(x). Watch the slope of the bottom graph and the height of the top graph at the same moment.

y = f(t), with the area from a to x shaded

y = A(x), the area so far, with its tangent line

Area function
x
Area so far A(x)
Slope of A at x
Height f(x)

The Fundamental Theorem of Calculus

Here is the key observation. Let A(x) be the area under f from a to x, and ask how fast that area grows as x moves. Moving the edge from x to x+h adds a thin strip of area about f(x)\cdot h. Divide by h and let h\to0: the rate at which area accumulates is the height of the curve.

Part 1
\frac{d}{dx}\int_a^x f(t)\,dt=f(x)

for continuous f. The area function is an antiderivative of f.

Part 2
\int_a^b f(x)\,dx=F(b)-F(a)

for any antiderivative F of f. Instead of a limit of Riemann sums, you find an antiderivative and subtract.

So the puzzle resolves like this. The slope of the area function is the height of the original function. Taking the area under a curve and then finding its slope brings you back to where you started, which is exactly what it means for two operations to be opposites. Two problems that look unrelated, finding tangent lines and finding areas, turn out to be one problem seen from two directions. That is why calculus is one subject rather than two.

Why can you use any antiderivative in Part 2? Two antiderivatives differ by a constant, and the constant cancels when you subtract.

f(x)An antiderivative F(x)
x^n,\ n\ne-1\frac{x^{n+1}}{n+1}
\frac1x\ln|x|
\sin x-\cos x
\cos x\sin x
e^xe^x

Solve it step by step

Use both parts of the theorem. Answers can be fractions, such as 14/3. The score counts steps you answer correctly on the first try.

Problems solved0
First-try steps0 of 0

    Common mistakes

    Subtracting in the wrong order

    It is F(b)-F(a), the upper limit first.

    Dropping a sign in −cos

    \int_0^{\pi}\sin x\,dx=[-\cos x]_0^{\pi}=(-(-1))-(-1)=2. Two minus signs are easy to lose.

    Forgetting the chain rule in Part 1

    If the upper limit is g(x) rather than x, then \frac{d}{dx}\int_a^{g(x)}f(t)\,dt=f(g(x))\,g'(x).