If $$\int_{0}^{9} f(x) dx = 4 $$
Find $$\int_{0}^{3} xf(x^2) dx $$
I'm not sure if I'm approaching the problem correctly, but I would first start off with doing a u-sub
Choose $u = x^2$
Then, $du/2x = dx$
$$\int_{0}^{3} f(u)du/2 $$
$$(1/2)\int_{0}^{3} f(u)du$$
Change the limits since $u = x^2$
$$(1/2)\int_{0}^{9} f(u)du $$
Since we know that $$\int_{0}^{9} f(x) dx = 4 $$
$(1/2)(4)$
$2$
On
Looks like we had similar first instincts.
I tried using the substitution $u = \sqrt{x}$ on the first integral
with $du = \frac{dx}{2\sqrt{x}}$
substituting for $u$ we get
$dx = 2udu$
When we rewrite the integral in terms of $u$ the limits go from 0 and 9 to 0 and 3 (Because $u = \sqrt{x}$).
And $f(x)$ is equivalent to $f(u^2)$
Thus, $$ 4 = \int_{0}^{9} f(x) dx = 2\int_{0}^{3} uf(u^2)du $$
So we can see that the first integral is equal to double the value of the second integral.
$ 4 / 2 = 2 $
I think we saw this question in the same place, maybe I'll see you sometime this fall. Go Maroons!
Your final answer is good, but $$\int\limits_{0}^{3} xf(x^2) \mathrm{d}x \neq\frac{1}{2}\int\limits_{0}^{3} f(u) \mathrm{d}u$$ You should change the limits of the definite integral at the same time you change the variables: $$\int\limits_{0}^{3} xf(x^2) \mathrm{d}x =\frac{1}{2}\int\limits_{0}^{9} f(u) \mathrm{d}u$$ Because the $$\int\limits_{a}^{b} f(x) \mathrm{d}x$$ means that you are integrating $f(x)$ from $x=a$ to $x=b$.
Or at least do something like this: $$\int\limits_{0}^{3} xf(x^2) \mathrm{d}x=\frac{1}{2}\int\limits_{x=0}^{x=3} f(u) \mathrm{d}u$$