How would you solve for x?

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

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- Dec 1st 2008, 09:39 PMPn0yS0ld13rSolve for upper limit of integral?
How would you solve for x?

$\displaystyle \displaystyle\int^x_0 (t^{3}-2t+3)\,dt = 4$ - Dec 1st 2008, 09:59 PMADARSH
(Hi)Hello,

$\displaystyle

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

$

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

$\displaystyle

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

$

so

$\displaystyle

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

$

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