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integrate t-3 from 0 to x^2
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\[\int\limits_{0}^{x^2} (t-3) dt\]
Expand the integrand and write it like this \[\Large \int_0^{x^2}tdt-3 \int_0^{x^2}1dt \]
then just apply regular exponential rules of integration. \[ \Large \int x^n dx= \frac{x^{n+1}}{n+1} \]
okay, thanks . so what would be the answer, i'm trying to check my work.
\[\Large \frac{t^2}{2}-3t \] Now just substitute the upper bounds \(t=x^2\)
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\[\Large \frac{x^4}{2}-3x^2 \]
so our teacher was wrong ... thanks, i was wondering why i got a different answer.
i already used that, i thought i might have entered something in wrong. thanks again
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