$$ \int_1^2 (2+3t)^{3/2} \ dt $$
I dont know how to solve this when $3/2$ is outside the brackets. If somebody could explain to me how to do this in the simplest way possible I would greatly appreciate it!
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Image not available. – orion Mar 29 '17 at 11:49
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The image is not available. – Rab Mar 29 '17 at 11:49
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my apologies, fixed now. – Jordan Mar 29 '17 at 13:10
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Do you know u- substitution? – Tyberius Mar 29 '17 at 13:13
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Set a change of variable $x = 2+3t, \ dx = 3 dt$. This gives the new bound $x\in [5,8]$ : $$ \int_1^2 (2+3t)^{3/2} \ dt = \frac13\int_5^8 x^{3/2}\ dx $$ Can you conclude using the well-known "$x^n$" primitive ?
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