Question

Evaluate the following integrals: A) Integral of (1 - x^7) dx from 0 to 1. B) Integral of (1 - x)^7 dx from 0 to 1.

Answer 1

Explanation:

We can evaluate this integral using integration by substitution, or u-substitution. We pick some part of the integrand to set equal to some variable (such as u, but any variable is an option). Good places to look at first include under a radical or in the denominator. This is not always the case, but it is in this one.

We can set u=1−x

Therefore,

du=−1dx
−du=dx

We can substitute these values into our integral. We get:

−∫1√udu

Which we can rewrite as:

−∫u−12du

Integrating, we get:

−2u12

From here you have two options on evaluating for the given limits of integration. You can either choose now to substitute 1−x back in for u and evaluate from 0 to 1, or you can change the limits of integration and evaluate with u. I will demonstrate both options.

Substituting 1−x back in for u,

−2(1−x)12
−2[(1−1)12−(1−0)12]
−2(−1)

Final answer: 2

Changing limits of integration:

u=1−x

u=1−(1)
u=0 (new upper limit)

u=1−0
u=1 (new lower limit)

Evaluating, we have

−2[(0)12−(1)12]
−2(−1)

Final answer: 2

Hope this helps!

Answer 2

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$<b>$
∫ dx /(x^7 - x) = ∫ dx / x(x^6 - 1) multiply top and bottom with x^5 = ∫ x^5 dx / x^6(x^6 - 1) let x^6 - 1 = u ==> x^6 = u + 1 6x^5 dx = du x^5 dx = du/6

$\huge\underline{\green {\textbf{}}}$

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