Definite integral as area and as a sum
A water tank is filled by a tap with a changing speed. The total water collected between two times is a definite integral of the flow rate.
The definite integral of f from a to b is the area under the curve between x = a and x = b (above the axis). It can be thought of as the limit of the sum of thin rectangles. In practice it is found by F(b) − F(a), where F is any antiderivative of f. No constant C is needed.
Formula: ∫ from a to b of f(x) dx = F(b) − F(a)
In the figure: The shaded part under y = x² between x = 0 and x = 3 has area ∫ from 0 to 3 of x² dx = 9. Point B is (3, 9).
Rule 1: Evaluate with an antiderivative — Integrate, put in the upper limit, then subtract the lower limit.
- Easy: ∫ from 1 to 3 of 2x dx = [x²] from 1 to 3 = 9 − 1 = 8.
- Medium: ∫ from 0 to π of sin x dx = [−cos x] from 0 to π = 1 + 1 = 2.
- Exam-type: ∫ from 1 to 4 of √x dx = [(2/3) x^(3/2)] from 1 to 4 = (2/3)(8 − 1) = 14/3.
Rule 2: Sum of thin strips — More strips give a better value.
- Easy: For ∫ from 0 to 2 of x dx, 4 strips of width 0.5 with right-end heights 0.5, 1, 1.5, 2 give 0.5 × 5 = 2.5, while the exact value is 2.
- Medium: ∫ from 0 to 1 of eˣ dx = e − 1.
- Exam-type: ∫ from −1 to 1 of x² dx = [x³/3] from −1 to 1 = 2/3.
The answer of a definite integral is a number, not a function. Subtract the lower value from the upper value.
Subtracting in the wrong order, or adding + C to a definite integral.
Top minus bottom.
Practice questions
- 1Evaluate ∫ from 1 to 3 of 2x dx.
- 2Evaluate ∫ from 0 to 2 of (x² + 1) dx.
- 3Evaluate ∫ from 0 to π/2 of cos x dx.
- 4Evaluate ∫ from 1 to e of dx/x.
- 5Evaluate ∫ from 0 to 1 of eˣ dx.
- 6Evaluate ∫ from 1 to 4 of √x dx.
- 7Using 4 equal strips and right-end heights, estimate ∫ from 0 to 2 of x dx.
- 8Evaluate ∫ from −1 to 1 of x² dx.
Show answers (try them yourself first)
- 19 − 1 = 8
- 2[x³/3 + x] from 0 to 2 = 8/3 + 2 = 14/3
- 3sin(π/2) − sin 0 = 1
- 4ln e − ln 1 = 1
- 5e − 1
- 614/3
- 70.5 × (0.5 + 1 + 1.5 + 2) = 2.5
- 82/3
Area function and the two fundamental theorems
The distance travelled up to time x grows as x grows. Its rate of growth is the speed at that moment. This is the whole idea of the fundamental theorem.
Let A(x) be the area under the curve y = f(t) from a fixed point a up to x. The first theorem says that A′(x) = f(x), so area function is an antiderivative of f. The second theorem says that the definite integral of f from a to b equals F(b) − F(a) for any antiderivative F.
Formula: A(x) = ∫ from a to x of f(t) dt, A′(x) = f(x)
Rule 1: First theorem — Differentiating the area function returns the original function.
- Easy: A(x) = ∫ from 0 to x of 2t dt = x², and A′(x) = 2x.
- Medium: d/dx ∫ from 1 to x of (t³ + 1) dt = x³ + 1.
- Exam-type: d/dx ∫ from 0 to x of cos t dt = cos x, which is ½ at x = π/3.
Rule 2: Second theorem — Use any antiderivative, the constant cancels.
- Easy: ∫ from 0 to 2 of 3t² dt = 8, so the area function A(x) = x³ has A(2) = 8.
- Medium: ∫ from 1 to 2 of (3x² − 2x) dx = [x³ − x²] from 1 to 2 = 4 − 0 = 4.
- Exam-type: ∫ from 0 to π/4 of sec²x dx = tan(π/4) − tan 0 = 1.
The upper limit is the variable in the area function, and the lower limit is fixed. The derivative returns f at the upper limit.
Thinking that d/dx of ∫ from a to x of f(t) dt equals f(a).
Area grows, its slope is the height.
Practice questions
- 1Find ∫ from 0 to x of 2t dt.
- 2Find d/dx ∫ from 1 to x of (t³ + 1) dt.
- 3If A(x) = ∫ from 0 to x of t dt, find A(3).
- 4Find d/dx ∫ from 2 to x of eᵗ dt.
- 5If A(x) = ∫ from 0 to x of 3t² dt, find A(2).
- 6Evaluate ∫ from 0 to π/4 of sec²x dx.
- 7Evaluate ∫ from 1 to 2 of (3x² − 2x) dx.
- 8Find the value of d/dx ∫ from 0 to x of cos t dt at x = π/3.
Show answers (try them yourself first)
- 1x²
- 2x³ + 1
- 3[t²/2] from 0 to 3 = 9/2
- 4eˣ
- 5A(x) = x³, so A(2) = 8
- 6tan(π/4) − tan 0 = 1
- 7[x³ − x²] from 1 to 2 = 4 − 0 = 4
- 8cos(π/3) = 1/2