# Modifying Limits of Definite Integral

This was a practice problem I was given:

Suppose that $\int_{-3}^8 f(x) dx = 5$. Use this information to determine the values of the constants $a, b, k$ that will satisfy the definite integral $\int_a^b kf(2x)dx = 5$.

Intuitively, I reasoned I would simply change $a$ and $b$ to account for the transformation of $f(x)$. I did so by setting $2x = -3$ and $2x = 8$, giving me $a = -1.5$ and $b=4$, which was correct. However, I was a bit stumped on $k$. I thought $k$ would be $1$ because the updated limits would account for any change in value of the integral. However, the answer is $k=2$.

I don't understand where how to obtain $k$, or why it is necessary. I have a feeling I am going about these types of problems wrong. Is there a more systematic way to approach these types of problems?

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Do you already know about basic $u$-substitution?

If so, let $u=2x \implies du=2\,dx \implies dx={1\over 2}du$. Then consider what the new limits of integration would be under this change of variables.

Also, this substitution changes the limits of integration like so: $x=-3\implies u=-6$, and $x=8\implies u=16$. Thus, \begin{align}\int_{a}^{b} kf(2x)\,dx=\int_{2a}^{2b} kf(u)\,{1\over 2}du&=5\\ {k\over 2}\int_{2a}^{2b} f(u)\,du&=5\\ \int_{2a}^{2b} f(u)\,du&={10\over k}. \end{align}

Now, since you know $2a=-3$, $2b=8$ yield the right-hand side above equal to $5$, then you know $a=-3/2$, $b=4$, $k=2$ work.

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Great explanation, I think I've got it now. Thank you. –  Paul Spinelli Feb 9 '13 at 22:32

Suppose that you knew an antiderivative $F(x)$ of $f(x)$, so that $F\,'(x)=f(x)$, and

$$\int_{-3}^8f(x)\,dx=F(8)-F(-3)\;.$$

Now you calculate

$$\int_a^bkf(2x)\,dx$$

by making the substitution $u=2x$, $du=2dx$:

$$\int_a^bkf(2x)\,dx=\frac{k}2\int_{2a}^{2b}f(u)\,du=\frac{k}2\big(F(2b)-F(2a)\big)\;.$$

Thus, you want $$\frac{k}2\big(F(2b)-F(2a)\big)=5\;.$$

Since we know almost nothing about $F$, it’s clear that we need to take $b=\frac82=4$ and $a=\frac{-3}2=-\frac32$, as you did. Then we’ll have $$\frac{k}2(5)=5\;,$$

which clearly requires that $k=2$.

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