Integrals
GradesPre-Calculus & CalculusIntegralsIntegration by Substitution & Area Between Curves

Integration by Substitution & Area Between Curves

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Integration by Substitution & Area Between Curves

Use u-substitution to evaluate integrals and compute areas between two curves.

Integration by substitution (u-substitution) is the integration counterpart of the chain rule. We replace a complicated expression with a simpler variable u, integrate, then substitute back. The area between two curves uses the definite integral of their difference.

∫f(g(x)) g′(x) dx→u=g(x)∫f(u) du\int f(g(x))\,g'(x)\,dx \xrightarrow{u=g(x)} \int f(u)\,du

U-Substitution Method

1Choose u = g(x) — usually the inner function of a composite.
2Compute du = g'(x) dx.
3Rewrite the entire integral in terms of u.
4Integrate with respect to u.
5Substitute g(x) back for u.
∫2x ex2 dx→u=x2∫eu du=eu+C=ex2+C\int 2x\,e^{x^2}\,dx \xrightarrow{u=x^2} \int e^u\,du = e^u + C = e^{x^2} + C
Area between curves: A=∫ab[f(x)−g(x)]dx(f(x)≥g(x))\text{Area between curves: } A = \int_a^b \left[f(x) - g(x)\right]dx \quad (f(x) \geq g(x))
📐Area Between Two Parabolas
Find the area between y = x and y = x². Intersections: x = x² → x = 0 and x = 1. A = ∫₀¹ (x − x²) dx = [x²/2 − x³/3]₀¹ = 1/2 − 1/3 = 1/6.
A=∫01(x−x2) dx=[x22−x33]01=12−13=16A = \int_0^1 (x - x^2)\,dx = \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^1 = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}
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Remember This!

For u-substitution with definite integrals, either convert the limits of integration to u-values (and skip back-substitution) or back-substitute to x first, then apply the original limits.

✏️ Try It!

Evaluate ∫ 3x²(x³ + 1)⁴ dx.

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