So my teacher thought it would fun to go through a whole chapter on integrals while i was down with bronchitis. uhh.. could someone help me with these problems
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Are you sure the first question is correct? Usually with substitution, part of the integrand looks like a derivative of another part, and this is not the case here. Was the numerator 12x^2? If so, use u = 12x^2. Find du (but remember that in the problem, you have 12x^2dx) and try to integrate. Let me know if you have more questions.
The second problem was correct as explained by filosophicphool, though I disagree with the solution to the third.
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I got an answer of 2xe^2x - e^2x for b,
with u = 4x dv = e^2x
du = 4 v = (1/2)e^2x
and then integral of udv = uv - integral of vdu.
can anyone confirm this? I haven't taken calculus in a year (took, BC calc junior year.)
then i got (1/4)3^-4 for problem c.
integrate x^-3dx as usual and get
(-1/4)x^-4 | 3 to ∞.
(-1/4)(∞)^-4 - (-1/4)(3^-4)
the first part goes to zero because the infinity is part of the denominator.
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The second problem was correct as explained by filosophicphool, though I disagree with the solution to the third.
∫1/(x^3) = (-1/2)x^(-2)|3 to ∞
(-1/2)(1/∞^2) + (1/2)(1/3^2) = 1/18
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http://tutorial.math.lamar.edu/
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