How would you solve for x? $\displaystyle \displaystyle\int^x_0 (t^{3}-2t+3)\,dt = 4$
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Hello, Originally Posted by Pn0yS0ld13r How would you solve for x? $\displaystyle \displaystyle\int^x_0 (t^{3}-2t+3)\,dt = 4$ $\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
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