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Calculus I Foundations (MATH 151) · Module 8 of 11

Techniques of Integration & Applications

How to actually find antiderivatives, and what integrals are used for beyond area.

In this module
  1. Basic Antiderivatives
  2. U-Substitution
  3. Area Between Curves
  4. Average Value of a Function
  5. Check yourself

Basic Antiderivatives

  • The power rule for integration: ∫xⁿ dx = xⁿ⁺¹/(n+1) + C for n ≠ −1; ∫1/x dx = ln|x| + C.
  • Every indefinite integral includes a "+C" (constant of integration), since the derivative of any constant is zero, so infinitely many antiderivatives differ only by a constant.
Worked example

Why does every indefinite integral include an arbitrary constant "+C"?

Solution. Because the derivative of a constant is always zero, so if F(x) is one antiderivative of f(x), then F(x)+C is also an antiderivative for any constant C, since they all have the same derivative.

U-Substitution

  • U-substitution reverses the chain rule: choose u to be an inner function, compute du, and rewrite the integral entirely in terms of u, ideally simplifying it to a basic form.
Worked example

Use u-substitution to evaluate ∫2x(x²+1)⁴ dx.

Solution. Let u = x²+1, so du = 2x dx. The integral becomes ∫u⁴ du = u⁵/5 + C = (x²+1)⁵/5 + C.

Area Between Curves

  • The area between two curves f(x) and g(x), where f(x) ≥ g(x), on [a,b] is ∫[a,b] [f(x) − g(x)] dx — the integral of the "top" function minus the "bottom" function.
Area between two curves
xyy = x (top)y = x² (bottom)
On [0, 1] the line y = x lies above y = x², so the area between them is ∫₀¹ (x − x²) dx = 1/2 − 1/3 = 1/6.
Worked example

Set up, without solving, the integral for the area between y = x² and y = x on the interval where x² ≤ x.

Solution. Since x ≥ x² on [0,1] where they intersect, the area is ∫[0,1] [x − x²] dx — the top function (x) minus the bottom function (x²).

Average Value of a Function

  • The average value of a continuous function f on [a,b] is (1/(b−a)) ∫[a,b] f(x) dx — the "height" of a rectangle with the same base and area as the region under the curve.
Worked example

How is the average value of a function over an interval related to the definite integral?

Solution. It's the definite integral of the function over that interval, divided by the length of the interval — effectively spreading the total accumulated area evenly across the interval's width.

Check yourself

2 of this module's 5 quiz questions. Pick your answers, then check them in the app, with an explanation for every question.

1. What is ∫xⁿ dx (for n ≠ −1)?

2. Why does every indefinite integral include a "+C"?

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