How do you evaluate the definite integral #int (-2x^2+3x+2)dx# from [0,2]?

Question

Aurora Alvarado · Accepted Answer

#int_0^2 (-2x^2+3x+2) \ dx = 14/3 # We use # \ int x^n \ dx=x^(n+1)/(n+1)# giving; #int_0^2 (-2x^2+3x+2) \ dx = [-2/3x^3+3/2x^2+2x]_0^2 # # " " = (-2/3(2^3)+3/2(2^2)+2(2)) - (0) # # " " = (-16/3 +6+4) - (0) # # " " = 14/3 #

Isabella Ackerman · Answer

Adam Adkins · Answer

undefined To evaluate the definite integral $ \int_{0}^{2} (-2x^2 + 3x + 2) \, dx $, you need to find the antiderivative of the integrand and then evaluate it at the upper limit of integration (2) and subtract the value of the antiderivative at the lower limit of integration (0).

The antiderivative of $ -2x^2 + 3x + 2 $ with respect to $ x $ is $ -\frac{2}{3}x^3 + \frac{3}{2}x^2 + 2x $.

Now, evaluate this antiderivative at the upper limit (2) and the lower limit (0) of integration:

$$ \left[ -\frac{2}{3}x^3 + \frac{3}{2}x^2 + 2x \right]_{0}^{2} $$

$$ = \left( -\frac{2}{3}(2)^3 + \frac{3}{2}(2)^2 + 2(2) \right) - \left( -\frac{2}{3}(0)^3 + \frac{3}{2}(0)^2 + 2(0) \right) $$

$$ = \left( -\frac{16}{3} + 6 + 4 \right) - \left( 0 + 0 + 0 \right) $$

$$ = \left( -\frac{16}{3} + 10 \right) $$

$$ = -\frac{16}{3} + \frac{30}{3} $$

$$ = \frac{14}{3} $$

So, the value of the definite integral $ \int_{0}^{2} (-2x^2 + 3x + 2) \, dx $ is $ \frac{14}{3} $.