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The correct answer is $7$. I figured out that if you substitute the inner argument of the second function, the end points become the same but flipped. That means we can just negate the integral and get the same endpoints. This is where I am now stuck. How do I get to the result with that knowledge?

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    $\begingroup$ Yes, and if you substitute $7-8x$ you need to scale the integral by factor $-8$ what give you $(-8)\cdot (-7) = 56$ again. You just forgot to replace the "dx" by the substituted one. $\endgroup$ – Gono Jan 21 at 15:49
  • $\begingroup$ @Gono Whoops! Dumb mistake on my end. Thanks :) $\endgroup$ – user737163 Jan 21 at 15:51
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First, let $u = 7 - 8x$. Then, $u(0) = 7$ and $u(19) = -145$. Also, $dx = -\frac{du}{8}$. Then, we wish to calculate:

$$\int_{0}^{19}f(7-8x)\ dx = \int_{u(0)}^{u(19)}-f(u)\ \frac{du}{8} = -\frac{1}{8}\int_{7}^{-145}f(u)\ du = \frac{1}{8}\int_{-145}^{7}f(u)\ du =\frac{56}{8} = \boxed{7}$$

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  • $\begingroup$ Yup! Thank you :) I forgot to substitute du in there as well. $\endgroup$ – user737163 Jan 21 at 15:52
  • $\begingroup$ Sorry, I had an error before. It's fixed now. $\endgroup$ – Joshua Wang Jan 21 at 15:59
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Substituting in the second argument you get

$$-\frac{1}{8}\int_7^{-145}f(y)dy=\frac{1}{8}\int_{-145}^{7}f(y)dy=\frac{56}{8}=7$$

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