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How do I evaluate #int5v (v^2 + 2)^2dv#?

Answer 1

I would use a Brute Force approach, expanding the square, multiplying by #5v# and integrating: #int5v(v^2+2)^2dv=int5v(v^4+4v^2+4)dv=# #int(5v^5+20v^3+20v)dv=5v^6/6+20v^4/4+20v^2/2+c=# #=5/6v^6+5v^4+10v^2+c#

Remember that #intx^ndx=x^(n+1)/(n+1)+c#

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Answer 2

Ezra Adams

To evaluate ∫(v^2 + 2)^2 dv from 0 to 5, you first expand the integrand, then integrate each term, and finally evaluate the integral from 0 to 5 by substituting the upper limit (5) and the lower limit (0) into the antiderivative and subtracting the results.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.