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Compute an integral of (2x-5)3


    \[ \int_0^3 (2x-5)^3 \, dx. \]

We could compute this directly, but to make the computation a bit simpler we use the fact that integrals are invariant under translation (Theorem 1.18 of Apostol). Translating to the left by \frac{5}{2} we have

    \begin{align*}  \int_0^3 (2x-5)^3 \, dx &= \int_{-\frac{5}{2}}^{\frac{1}{2}} \left( 2 \left( x + \frac{5}{2} \right) - 5 \right)^3 \, dx \\[9pt]  &= \int_{-\frac{5}{2}}^{\frac{1}{2}} \left( 2x + 5 - 5 \right)^3 \, dx \\[9pt]  &= \int_{-\frac{5}{2}}^{\frac{1}{2}} (2x)^3 \, dx \\[9pt]  &= \left. 8 \frac{x^4}{4} \right|_{-\frac{5}{2}}^{\frac{1}{2}} \\[9pt]  &= \frac{1}{8} - \frac{625}{8} \\[9pt]  &= -78. \end{align*}


    • Rori says:

      Hi, I updated the post with some extra details. The translation is to the left by \frac{5}{2} (so the interval of integration moves to the left by that amount) and then we replace x with \left( x + \frac{5}{2} \right). Of course, cubing that expression and working it out directly should get to the same answer. Let me know if the extra details are clear.

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