I have an integral from t=0 to 40 of (t/40)^(a+1)dt. I can get have way to the known answer of 40/(a+2) by substitution to using u=(t/40) and du=(1/40) which resolve to 40 times the integral over 0 to 1 of u^(a+1)du. I am simply stumped now how to get to 40/(a+2). What rules am I not seeing? Cheers A.

Question

I have an integral from t=0 to 40 of (t/40)^(a+1)dt.  I can get have way to the known answer of 40/(a+2) by substitution to using u=(t/40) and du=(1/40) which resolve to 40 times the integral over 0 to 1 of u^(a+1)du.  I am simply stumped now how to get to 40/(a+2).  What rules am I not seeing? Cheers A.

hartnn · Answer

you know the formula ., 
$\Large \int x^n dx = \dfrac{x^{n+1}}{n+1}+c$

hartnn · Answer

using that, 
$\Large 40\int u^{a+1}du=40 \dfrac{u^{a+2}}{a+2}$

hartnn · Answer

then apply your limits of u=0 and u=1

anonymous · Answer

Okay, thanks.  I will have a scratch at that and see how it gels.  I am doing my masters and unfortunately haven't used calculus in 20+ years.  Fun but slow.

anonymous · Answer

DOH! One raised to any power is one.  So close but sooooo rusty.  Thanks hartnn.

hartnn · Answer

welcome ^_^
oh, and since you are new here, 
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