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

Question

Cameron Alexander · Accepted Answer

#:. int_-1^2(-x^2+x+2)dx=9/2=4.5#

By the Fundamental Theorem of Calculus, we know that,

#intf(x)dx=F(x)+c rArr int_a^bf(x)dx=[F(x)]_a^b=F(b)-F(a)#

#:. int_-1^2(-x^2+x+2)dx=[-x^3/3+x^2/2+2x]_-1^2#

#=(-8/3+2+4)-{-(-1)/3+1/2-2}#

#=(-8/3-1/3)+6+2-1/2#

#=-3+8-1/2#

#:. int_-1^2(-x^2+x+2)dx=9/2=4.5#

Adam Adkins · Answer

undefined To evaluate the definite integral $\int_{-1}^{2} (-x^2+x+2) \, dx$, you need to find the antiderivative of the function $-x^2+x+2$, and then evaluate it at the upper and lower limits of integration, and finally subtract the value at the lower limit from the value at the upper limit.

First, find the antiderivative:
$$\int (-x^2+x+2) \, dx = -\frac{1}{3}x^3 + \frac{1}{2}x^2 + 2x + C$$

Now evaluate this antiderivative at the upper and lower limits of integration:
$$F(2) - F(-1) = \left(-\frac{1}{3}(2)^3 + \frac{1}{2}(2)^2 + 2(2)\right) - \left(-\frac{1}{3}(-1)^3 + \frac{1}{2}(-1)^2 + 2(-1)\right)$$

$$= \left(-\frac{8}{3} + 2 + 4\right) - \left(\frac{1}{3} + \frac{1}{2} - 2\right)$$

$$= \left(-\frac{8}{3} + 6\right) - \left(\frac{1}{6} - 2\right)$$

$$= \left(\frac{10}{3}\right) - \left(-\frac{11}{6}\right)$$

$$= \frac{10}{3} + \frac{11}{6}$$

$$= \frac{20}{6} + \frac{11}{6}$$

$$= \frac{31}{6}$$

So, $\int_{-1}^{2} (-x^2+x+2) \, dx = \frac{31}{6}$.