## anonymous 5 years ago integral of (3x+1) dx? Upper limit =1. lower limit=0

1. anonymous

$\int\limits_{0}^{1} (3x+1)dx$ please explain. Im getting the answer 4. But im pretty sure its not correct...

2. anonymous

(3x^2)/2 + x | (3(1)/2 + 1) -0 (because 3x0/2 +0 is obviously just 0) = 2 & 1/2

3. anonymous

woops. sorry, the equation is actually :: $\int\limits_{0}^{1} (3x+1)^{2}$. The answer is 7 but im getting 4

4. anonymous

first, foil. (3x+1)^2= 9x^2+6x+1 dx = 3x^2+3x+x | (3+3+1) - 0 = 7

5. anonymous

I expanded (3x +1)^2 to $(3x+1) (3x+1)$ Then factored it. Then integrated it ..

6. anonymous

okay, thank you!