Math · Calculus I · Worked example

Change the limits in a definite substitution

Evaluate ∫₀² x(x² + 1)³ dx.

\int_{0}^{2} x {(x^{2} + 1)}^{3} d x

Choose u and find du

Let u = x² + 1. Then du = 2x dx, so x dx = du/2.

u = x^{2} + 1, x d x = \frac{d u}{2}

Convert the limits

When x = 0, u = 0² + 1 = 1. When x = 2, u = 2² + 1 = 5.

x = 0 \to u = 1, x = 2 \to u = 5

Integrate in u

The integral becomes ∫₁⁵ u³/2 du. An antiderivative of u³/2 is u⁴/8.

\int_{1}^{5} \frac{u^{3}}{2} d u = \frac{5^{4} - 1^{4}}{8} = 78

Stay in u

The limits are already u-values, so there is no need to go back to x. As a check, the x-antiderivative (x² + 1)⁴/8 gives 625/8 − 1/8 = 78 with the original limits.

Result

∫₀² x(x² + 1)³ dx = 78.

Your turn

Evaluate ∫₀¹ 2x(x² + 1)² dx.

Show the answer and explanation

7/3.

With u = x² + 1, du = 2x dx and the limits become u = 1 and u = 2. Then ∫₁² u² du = (2³ − 1³)/3 = 7/3.

\int_{1}^{2} u^{2} d u = \frac{2^{3} - 1^{3}}{3} = \frac{7}{3}

Keep exploring

Evaluate u⁴/8 at the old limits 0 and 2 instead and you get 2, not 78. That is the mix-up between x-limits and u-limits.

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