# Math Help - Solve for upper limit of integral?

1. ## Solve for upper limit of integral?

How would you solve for x?

$\displaystyle\int^x_0 (t^{3}-2t+3)\,dt = 4$

2. Hello,
Originally Posted by Pn0yS0ld13r
How would you solve for x?

$\displaystyle\int^x_0 (t^{3}-2t+3)\,dt = 4$
$
I= \int^x_0 {t^3-2t+3}
$

$I= \int^x_0{t^3dt} - \int^x_0{2tdt}+ \int^x_0{3dt}$

$
I=[\frac{t^4}{4}-{t}^2+3t]^x_0 = 4
$

so
$
\frac{x^4}{4}-{x}^2+3x-4=0

$

That's the equation You will get the answer as -3.091