Integration · Technique guide
Integrals and Areas in Your Head
By Robert Hansson, Kung Fu Math
Integration sounds like the most advanced topic in school maths, but many integrals can be solved in your head. An integral is an area, and many areas are triangles, rectangles or trapezoids in disguise. The rest often yield to one rule: the power rule in reverse.
These techniques will not replace a full course in calculus. They will, however, let you check answers instantly, spot when a result is impossible and solve simple exam questions without writing a single line.
01
Reverse the power rule
Differentiating xⁿ gives n·xⁿ⁻¹. Integration goes the other way: raise the power by one, then divide by the new power. So ∫xⁿ dx = xⁿ⁺¹/(n + 1) + C.
Check every antiderivative by differentiating it in your head. If you get back the original function, you are right.
Worked example
∫₀² 3x² dx
- → Antiderivative of 3x² is x³
- → Evaluate at 2: 8
- → Evaluate at 0: 0
- → 8 − 0 = 8
Answer: 8
Practice: Try ∫₀³ 2x dx, ∫₁² 4x³ dx and ∫₀¹ (6x² + 1) dx.
02
See the triangle
The graph of a straight line through the origin forms a triangle with the x-axis. The integral is simply base × height ÷ 2.
For ∫₀⁶ x dx the triangle has base 6 and height 6, so the area is 18. No antiderivative needed.
Worked example
∫₀⁴ 3x dx
- → Base: 4
- → Height at x = 4: 3 · 4 = 12
- → Area: 4 · 12 ÷ 2 = 24
Answer: 24
Practice: Try ∫₀⁵ 2x dx, ∫₀¹⁰ x/2 dx and ∫₀³ 4x dx.
03
Use the trapezoid for straight lines
For any straight line over an interval, the area is a trapezoid: the average of the two end heights times the width. ∫ₐᵇ (kx + m) dx = (b − a) · (f(a) + f(b)) ÷ 2.
This is exact for linear functions and a good estimate for gently curved ones.
Worked example
∫₁⁵ (2x + 1) dx
- → Heights: f(1) = 3 and f(5) = 11
- → Average height: (3 + 11) ÷ 2 = 7
- → Width: 5 − 1 = 4
- → Area: 4 · 7 = 28
Answer: 28
Practice: Try ∫₂⁶ (x + 3) dx, ∫₀⁴ (5 − x) dx and ∫₁³ (3x − 1) dx.
04
Let symmetry do the work
An odd function, such as x, x³ or sin x, is mirrored with a sign change around the origin. Over a symmetric interval from −a to a its positive and negative areas cancel, so the integral is 0.
An even function, such as x² or cos x, is mirrored without sign change. Its integral from −a to a is twice the integral from 0 to a.
Worked example
∫₋₃³ (x³ + 2) dx
- → x³ is odd: its part is 0
- → 2 is a constant: a rectangle of width 6 and height 2
- → Area: 6 · 2 = 12
Answer: 12
Practice: Try ∫₋₂² (x⁵ − x) dx, ∫₋₁¹ 3x² dx and ∫₋π^π sin x dx.
05
Estimate a definite integral
When an exact answer is hard, estimate. The midpoint rule takes the function value in the middle of the interval and multiplies by the width. For curves that bend gently, it is surprisingly close.
Estimating is a superb way to check a calculated answer. If your exact result is far from the estimate, look for an error.
Worked example
∫₀² x² dx (estimate)
- → Midpoint: x = 1, value 1² = 1
- → Width: 2, estimate 2 · 1 = 2
- → Exact value: 8/3 ≈ 2.67
- → Two midpoints (0.5 and 1.5) give 0.25 + 2.25 = 2.5, closer still
Answer: ≈ 2.67
Practice: Estimate ∫₁³ x² dx and ∫₀⁴ √x dx, then compare with the exact values.
Common mistakes
- Dividing by the old power instead of the new one: ∫x² dx is x³/3, not x³/2.
- Forgetting that area below the x-axis counts as negative in a definite integral.
- Using odd-function symmetry on an interval that is not symmetric around 0.
- Forgetting + C in an indefinite integral.
Summary
- Raise the power by one and divide by the new power.
- Straight lines give triangles and trapezoids: use geometry.
- Odd functions over −a to a give 0; even functions give twice the half.
- Estimate with the midpoint rule to check your answer.
Into the dojo
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