g(x)=∫_{-7}^{x}(t - t²)dt\n g(10)=

Explanation:

Step1: Apply the fundamental theorem of calculus

If $g(x)=\int_{a}^{x}f(t)dt$, then $g'(x) = f(x)$. Here $a=-7$ and $f(t)=t - t^{2}$, so $g'(x)=x - x^{2}$.

Step2: Evaluate $g'(x)$ at $x = 10$

Substitute $x = 10$ into $g'(x)$. We get $g'(10)=10-10^{2}$. $g'(10)=10 - 100=-90$.

Answer:

$-90$