Question

integral of (3x+1) dx? Upper limit =1. lower limit=0

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

okay, thank you!