Answer: [tex]-\frac{\sqrt{35}}{2}[/tex]
In terms of typing this in on a keyboard, you could say -sqrt(35)/2
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Work Shown:
We'll use u-substitution here
Let u = x^2, so du/dx = 2x which rearranges to du = 2x*dx and we can say xdx = du/2
Because we are changing from x^2 to u, this means we need to change the limits of integration
If x = -2, then u = x^2 = (-2)^2 = 4
If x = 0, then u = x^2 = 0^2 = 0
Despite 4 being larger than 0, we still place 4 at the bottom limit and we'll just swap them later.
This means we have the following steps
[tex]\displaystyle \int_{-2}^{0}xg(x^2)\sqrt{7}dx\\\\\\\displaystyle \sqrt{7}\int_{-2}^{0}xg(x^2)dx\\\\\\\displaystyle \sqrt{7}\int_{-2}^{0}g(x^2)xdx\\\\\\\displaystyle \sqrt{7}\int_{4}^{0}g(u)\frac{du}{2}\\\\\\\displaystyle -\sqrt{7}\int_{0}^{4}g(u)\frac{du}{2}\\\\\\\displaystyle -\sqrt{7}*\frac{1}{2}\int_{0}^{4}g(u)du\\\\\\\displaystyle -\frac{\sqrt{7}}{2}\sqrt{5}\\\\\\\displaystyle -\frac{\sqrt{7}*\sqrt{5}}{2}\\\\\\\displaystyle -\frac{\sqrt{7*5}}{2}\\\\\\\displaystyle -\frac{\sqrt{35}}{2}\\\\\\[/tex]
Notice when I swapped the limits of integration (4 and 0 to 0 and 4) I placed a negative sign out front.
The rule is [tex]\displaystyle \int_{a}^{b}f(x)dx = -\int_{b}^{a}f(x)dx[/tex]
