Compute
![\[ \int_0^3 (2x-5)^3 \, dx. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-761886ab6e08e92997e55d2cd686dd89_l3.png "Rendered by QuickLaTeX.com")
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 
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*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-9a1c59097cf477c27bf746f88f7bee8e_l3.png "Rendered by QuickLaTeX.com")
Compute
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 we have
Thanks. Got it now. Nice blog by the way.
Hi, I’m also working through Apostol. I don’t see how you used the translation invariance in here.
Hi, I updated the post with some extra details. The translation is to the left by
(so the interval of integration moves to the left by that amount) and then we replace 
with 
. 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.