TL;DR:
Integrals and derivatives undo each other. The rate of change of accumulated area is the height of the curve.
To compute
∫ₐᵇ f(x) dx, find an antiderivative F and calculateF(b) − F(a).For
f(x) = 2x, the antiderivative isx², so the area from 0 to 2 is4 − 0 = 4.
Two ideas have been running side by side: the derivative (slope) and the integral (area). It would be a surprise if they had nothing to do with each other. The fundamental theorem of calculus says they are two sides of the same coin, and it makes computing areas far easier than adding up rectangles.
Accumulating area
Imagine filling a bath. The height of the graph at time t is the rate the water flows in. The amount of water in the bath at time x is the accumulated area under the graph from 0 to x. Call it A(x). If the tap is running faster at that moment, the bath fills faster. In other words, the rate at which A grows is the height of the graph. That is the first half of the theorem: A′(x) = f(x).
Worked example: the integral of 2x from 0 to 2 equals 4
Draw y = 2x. From 0 to 2 it is a straight line rising from 0 to 4, so the area is a triangle: ½ × 2 × 4 = 4.
Now do it with the theorem. We need a function whose derivative is 2x. By the power rule, the derivative of x² is 2x, so F(x) = x² is an antiderivative of 2x.
Evaluate F at the top limit:
F(2) = 4.Evaluate F at the bottom limit:
F(0) = 0.Subtract:
F(2) − F(0) = 4 − 0 = 4.
The two methods give 4. That agreement is the theorem in action: an area you can see equals a subtraction you can compute.
Checking the accumulation idea
Define A(x) = ∫₀ˣ 2t dt. By the triangle rule, the area up to x is ½ × x × 2x = x². So A(x) = x², and its derivative is A′(x) = 2x, the height of the graph. That is the first half of the theorem in a concrete case.
A second computation: 3x² from 1 to 3
An antiderivative of 3x² is x³, since the derivative of x³ is 3x². Then the integral from 1 to 3 is F(3) − F(1) = 27 − 1 = 26. Check with a rough sum: two rectangles of width 1 with heights at the midpoints 1.5 and 2.5 give 3(2.25) + 3(6.25) = 6.75 + 18.75 = 25.5. That is close to 26, as it should be for a rough sum.
Try one yourself: x² + 1 from 0 to 3
An antiderivative of x² + 1 is F(x) = x³/3 + x. Then F(3) = 9 + 3 = 12 and F(0) = 0, so the area is 12. Check with midpoint rectangles of width 1, heights at 0.5, 1.5 and 2.5: (1.25 + 3.25 + 7.25) × 1 = 11.75. That is close to 12, and a finer sum would be closer.
Now a case below the axis: the integral of x − 2 from 0 to 2. An antiderivative is x²/2 − 2x, so the value is (2 − 4) − 0 = −2. Picture it: the line runs from −2 up to 0, a triangle of base 2 and height 2 below the axis. Its area is ½ × 2 × 2 = 2, counted as negative. The theorem handles the sign automatically.
Reading it as distance and speed
Let s(t) be position and v(t) = s′(t) be speed. The theorem says the integral of speed from t = a to t = b is s(b) − s(a): the change in position. If v(t) = 2t m/s, then s(t) = t² works, and from 0 to 2 s the object moves 4 − 0 = 4 m. The same 4 as the area, the same 4 as the triangle. Slope of position gives speed, and area under speed gives back the change in position.
Why the constant does not matter
Many functions have the same derivative. x², x² + 5 and x² − 100 all have derivative 2x. Try F(x) = x² + 5 on our example: F(2) − F(0) = 9 − 5 = 4. The constant cancels in the subtraction, so any antiderivative will do.
How to check your answer
Differentiate your antiderivative. You must get back the original function.
Compare with geometry. For lines, a triangle or trapezoid gives the area.
Order of subtraction. Top limit minus bottom limit.
Sign. If the curve is above the axis the answer is positive.
Common mistakes
Subtracting in the wrong order.
Forgetting to evaluate at both limits.
Using the derivative instead of an antiderivative.
Dropping the exponent rule: the antiderivative of
x²isx³/3, notx³. Differentiate to test:3x²/3 = x².Adding a constant and then forgetting it cancels, or carrying it into the answer.
Where you will use it next
The theorem relies on the idea of a limit, treated in limits: what happens near a point, and rests on what an integral adds up. It lets you find exact distances, volumes and averages without adding rectangles.
Frequently asked questions
Why is it called fundamental?
Because it links the two main operations of calculus, taking slopes and finding areas, and turns the second into the reverse of the first.
Can every function be integrated this way?
Every continuous function has an antiderivative, but it is not always a simple formula. In those cases, numerical sums are used.
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