Answer: [tex]-0.5(x^2+1)^5+2.5[/tex]
This is the same as writing -0.5(x^2+1)^5 + 2.5
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Explanation:
We take the integral to find the original f(x) function. Another term for "integral" is "antiderivative", which basically is the inverse of a derivative. There's more to the story, but this is the basic idea.
Apply a u-substitution
u = x^2+1
du/dx = 2x
du = 2xdx
xdx = du/2 = 0.5du
So,
[tex]\displaystyle \int f ' (x) dx\\\\\\\displaystyle \int -5x(x^2+1)^4 dx\\\\\\\displaystyle -5\int (x^2+1)^4 xdx\\\\\\\displaystyle -5\int u^4(0.5du)\\\\\\\displaystyle -5*0.5\int u^4du\\\\\\\displaystyle -2.5\int u^4du\\\\\\-2.5\left(\frac{1}{4+1}u^{4+1}+C\right) \ \text{ ... don't forget about the +C}\\\\\\-2.5\left(\frac{1}{5}u^{5}+C\right)\\\\\\-2.5\left(\frac{1}{5}(x^2+1)^{5}+C\right)\\\\\\-0.5(x^2+1)^{5}-2.5C\\\\\\[/tex]
This means that
[tex]f(x) = -0.5(x^2+1)^{5}-2.5C\\\\\\[/tex]
To determine the constant C, we plug in x = 0 and replace f(x) with 2.
This is directly from the fact that f(0) = 2.
[tex]f(x) = -0.5(x^2+1)^{5}-2.5C\\\\f(0) = -0.5(0^2+1)^{5}-2.5C\\\\2 = -0.5-2.5C\\\\-2.5C = 2+0.5\\\\-2.5C = 2.5\\\\C = 2.5/(-2.5)\\\\C = -1[/tex]
Therefore,
[tex]f(x) = -0.5(x^2+1)^{5}-2.5C\\\\\\f(x) = -0.5(x^2+1)^{5}-2.5(-1)\\\\\\f(x) = -0.5(x^2+1)^{5}+2.5\\\\\\[/tex]
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Checking the answer:
Plug in x = 0. You should find that f(0) = 2. I'll let you do these steps.
Apply the derivative and use the chain rule to find:
[tex]f(x) = -0.5(x^2+1)^5+2.5\\\\f'(x) = 5*(-0.5)(x^2+1)^{5-1}*\frac{d}{dx}(x^2+1)\\\\f'(x) = 5*(-0.5)(x^2+1)^{4}*(2x)\\\\f'(x) = -5x(x^2+1)^{4}\\\\[/tex]
The answer is fully confirmed. Another way you can confirm the answer is to use free tools like GeoGebra or WolframAlpha. I show an example of this in the screenshot below.
