How do you evaluate the definite integral #int (x-2) dx# from #[-1,0]#?

Question

Cameron Alexander · Accepted Answer

#-5/2#

#int_-1^0(x-2)dx#

Using the power formula for integral computation, increase the exponent on the variable and divide each term by its new exponent.

#[x^2/2-(2x^1)/1]_-1^0#

#[x^2/2-2x]_-1^0#

Now plug in the upper and lower bounds of integration as values for x and subtract.

#[((0)^2/2-2(0))-((-1)^2/2-2(-1))]#

#[(0-0)-(1/2+2)]=-5/2#

Samantha Adkison · Answer

undefined To evaluate the definite integral $ \int_{-1}^0 (x - 2) \, dx $, you need to find the antiderivative of $ x - 2 $ and then evaluate it at the upper and lower limits of integration and take their difference.

The antiderivative of $ x - 2 $ with respect to $ x $ is $ \frac{x^2}{2} - 2x $.

Evaluate this antiderivative at the upper limit of integration (0) and subtract the result when evaluated at the lower limit of integration (-1):

$$ \left[ \frac{x^2}{2} - 2x \right]_{-1}^0 $$

$$ = \left( \frac{0^2}{2} - 2(0) \right) - \left( \frac{(-1)^2}{2} - 2(-1) \right) $$

$$ = (0 - 0) - \left( \frac{1}{2} + 2 \right) $$

$$ = -\frac{1}{2} - 2 $$

$$ = -\frac{5}{2} $$

So, the value of the definite integral $ \int_{-1}^0 (x - 2) \, dx $ is $ -\frac{5}{2} $.