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Starting with F(x)=∫(-2 to x) 3t^2(cos(t^3)+2)dt, use the substitution u(t)=t^3 to rewrite the definite integral.
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\(\large F(x) = \int \limits_{-2}^x 3t^2 (\cos(t^3) + 2)dt\)
substitute \(\large u = t^3\)
\(\large du = 3t^2 dt\)
next work the bounds, do u knw how to figure out the new bounds for \(u\) ?
thank you for replying! I'm not completely sure, can you please help?
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the original bounds are (-2, x) right ?
when u substitute u = t^3, the bounds change to : ((-2)^3, x^3) (-8, x^3)
so the integral becomes :- \(\large F(x) = \int \limits_{-8}^{x^3} (\cos(u) + 2) du\)
see if that makes some sense.. .
aha! I didn't realize how simple that was. thank you!
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:) u wlc !
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