Answer: [tex]F'(x)=-6x^3+x-1[/tex]
Explanation:
The second fundamental theorem of calculus (2nd FTC) says the following. In words, the derivative is equal to the function evaluated at the upper limit times the derivative of the upper limit minus the function evaluated at the lower limit times the derivative of the lower limit.
[tex]\displaystyle \int\limits^b_a {f(t)} \, dx =f(x)|^b_a=f(b)b'-f(a)a'[/tex]
We will apply this theorem to evaluate the given integral.
[tex]\displaystyle F(x)\int\limits^{x}_{-x^2} {(3t-1)} \, dx \\F'(x)=f(3t-1)|^x_{-x^2}\\ F'(x)=(3(x)-1)(1)-(3(-x^2)-1)(-2x)[/tex]
Next, we will simplify by multiplying and subtracting.
[tex]F'(x)=(3x-1)(1)-(-3x^2-1)(-2x)\\F'(x)=3x-1-(6x^3+2x)\\F'(x)=-6x^3+x-1[/tex]
