How do you evaluate the definite integral #int (x+3)(x-1)dx# from [1,3]?

Question

Charles Arocha · Accepted Answer

#35/3#

First step is to distribute the brackets.

#rArrint_1^3(x+3)(x-1)dx#

#=int_1^3(x^2+2x-3)dx#

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);n≠-1)color(white)(2/2)|)))#

#=[1/3x^3+x^2-3x]_1^3#

#color(blue)"Note: "[F(x)]_a^b=[F(b)]-[F(a)]#

#rArr(9+9-9)-(1/3+1-3)=35/3#

Isabella Ackerman · Answer

undefined To evaluate the definite integral $\int_{1}^{3} (x+3)(x-1) \, dx$, you first need to expand the integrand and then integrate it term by term. After integrating, you evaluate the resulting expression at the upper limit of integration (in this case, 3) and subtract the value obtained by evaluating the expression at the lower limit of integration (in this case, 1).

Here are the steps:

1. Expand the integrand:
   $(x+3)(x-1) = x^2 + 2x - 3$.

2. Integrate term by term:
   $\int (x^2 + 2x - 3) \, dx = \frac{1}{3}x^3 + x^2 - 3x + C$.

3. Evaluate the integrated expression at the upper limit (3) and lower limit (1):
   $= \left[\frac{1}{3}(3)^3 + (3)^2 - 3(3)\right] - \left[\frac{1}{3}(1)^3 + (1)^2 - 3(1)\right]$.

4. Calculate the values:
   $= \left[\frac{27}{3} + 9 - 9\right] - \left[\frac{1}{3} + 1 - 3\right]$
   $= (9 + 9 - 9) - (\frac{1}{3} + 1 - 3)$
   $= 9 - (\frac{1}{3} - 2)$
   $= 9 - (-\frac{5}{3})$
   $= 9 + \frac{5}{3}$
   $= \frac{27 + 5}{3}$
   $= \frac{32}{3}$.

Therefore, the value of the definite integral $\int_{1}^{3} (x+3)(x-1) \, dx$ is $\frac{32}{3}$.