How do you find the antiderivative of int (2x^3-1)(x^4-2x)^6dx∫(2x3−1)(x4−2x)6dx from [-1,1][−1,1]?

1 Answer
mason m
Nov 9, 2016

int_(-1)^1(2x^3-1)(x^4-2x)^6dx=(-1094)/7∫1−1(2x3−1)(x4−2x)6dx=−10947

Explanation:

int_(-1)^1(2x^3-1)(x^4-2x)^6dx∫1−1(2x3−1)(x4−2x)6dx

We will apply the substitution u=x^4-2xu=x4−2x. Differentiating this reveals that du=(4x^3-2)dxdu=(4x3−2)dx. Note that the (2x^3-1)(2x3−1) term is exactly half this, so we can modify the integral as follows:

=1/2int_(-1)^1(x^4-2x)^6(4x^3-2)dx=12∫1−1(x4−2x)6(4x3−2)dx

Now we can substitute in our uu and dudu values. However, we also should remember to change our bounds! Take -1−1 and 11 and plug them into our u=x^4-2xu=x4−2x term. That is, the bound of -1−1 becomes (-1)^4-2(-1)=3(−1)4−2(−1)=3 and the bound of 11 becomes 1^4-2(1)=-114−2(1)=−1. Thus:

=1/2int_3^(-1)u^6du=1/2[u^7/7]_3^(-1)=1/2[(-1)^7/7-3^7/7]=12∫−13u6du=12[u77]−13=12[(−1)77−377]

=1/2[(-2188)/7]=(-1094)/7=12[−21887]=−10947

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