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Back to Integral Calculus

Integration by Substitution

A change of variable stretches the axis. Every thin strip gets wider by u′(x) and shorter by the same factor — so the area, and the integral, come through unchanged.

Step 1 of 24

In x: g(x) = 2x cos(x²)

Speed

∫ from 0 to 1.5 of 2x·cos(x²) dx. The integrand is awkward, but it contains x² and, beside it, 2x — the derivative of x². That pairing is the signal for a substitution.

\int_0^{1.5} 2x\cos(x^2)\,dx \;\overset{u = x^2}{=}\; \int_{0}^{2.25} \cos u\,du = 0.7781

\text{direct (Simpson, in } x\text{)}: 0.7781 \qquad \text{after substituting}: 0.7781

Integration by Substitution

A change of variable stretches the axis. Every thin strip gets wider by u′(x) and shorter by the same factor — so the area, and the integral, come through unchanged.

Step 1 of 24

In x: g(x) = 2x cos(x²)

Speed

∫ from 0 to 1.5 of 2x·cos(x²) dx. The integrand is awkward, but it contains x² and, beside it, 2x — the derivative of x². That pairing is the signal for a substitution.

\int_0^{1.5} 2x\cos(x^2)\,dx \;\overset{u = x^2}{=}\; \int_{0}^{2.25} \cos u\,du = 0.7781

\text{direct (Simpson, in } x\text{)}: 0.7781 \qquad \text{after substituting}: 0.7781

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Integration by Substitution

See also

Integration by Substitution

See also