# Solve for upper limit of integral?

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• Dec 1st 2008, 10:39 PM
Pn0yS0ld13r
Solve for upper limit of integral?
How would you solve for x?

$\displaystyle\int^x_0 (t^{3}-2t+3)\,dt = 4$
• Dec 1st 2008, 10:59 PM
ADARSH
(Hi)Hello,
Quote:

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