Question

for problems 7 – 10, leave as an exact value {no rounding}.

1. find the definite integral from x = -4 up to x = 2 for the following:

$f(x) = 3x^2 + 8x^3 - 20x - 4$

Explanation:

Step1: Find the antiderivative

$\int (3x^2 + 8x^3 - 20x - 4)dx = x^3 + 2x^4 - 10x^2 - 4x + C$

Step2: Apply Fundamental Theorem of Calculus

Evaluate at $x=2$ and $x=-4$, then subtract:
First, at $x=2$:
$(2)^3 + 2(2)^4 - 10(2)^2 - 4(2) = 8 + 32 - 40 - 8 = -8$
At $x=-4$:
$(-4)^3 + 2(-4)^4 - 10(-4)^2 - 4(-4) = -64 + 512 - 160 + 16 = 304$
Now subtract: $(-8) - 304$

Step3: Calculate the final result

$-8 - 304 = -312$

Answer:

$-312$