How do you evaluate the definite integral #int s^3-3s^2+2# from #[-2,2]#?

Question

Isabella Ackerman · Accepted Answer

#-8#

Integrate each term using the #color(blue)"power rule for integration"#

#color(orange)"Reminder " color(red)(bar(ul(|color(white)(2/2)color(black)(int(ax^n)=a/(n+1)x^(n+1))color(white)(2/2)|)))#

#rArrint_(-2)^2(s^3-3s^2+2)ds#

#=[1/4s^4-s^3+2s]_(-2)^2#

Evaluate for s = 2 and subtract the evaluation for s = - 2

#=(4-8+4)-(4+8-4)=0-8=-8#

Samantha Adkison · Answer

undefined To evaluate the definite integral $\int_{-2}^{2} (s^3 - 3s^2 + 2) \, ds$, you first integrate the function $s^3 - 3s^2 + 2$ with respect to $s$ and then evaluate it at the upper and lower limits of integration (in this case, -2 and 2).

$\int_{-2}^{2} (s^3 - 3s^2 + 2) \, ds = \left[\frac{1}{4}s^4 - s^3 + 2s\right]_{-2}^{2}$

Now, substitute the upper limit of integration (2) into the integrated function:

$\left[\frac{1}{4}(2)^4 - (2)^3 + 2(2)\right]$

Then, substitute the lower limit of integration (-2) into the integrated function:

$-\left[\frac{1}{4}(-2)^4 - (-2)^3 + 2(-2)\right]$

Now, calculate the values:

$= \left[\frac{1}{4}(16) - 8 + 4\right] - \left[\frac{1}{4}(16) + 8 - 4\right]$

$= \left[4 - 8 + 4\right] - \left[4 + 8 + 4\right]$

$= 0 - 16$

$= -16$

So, the value of the definite integral is $-16$.