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Chapter 7: Problem 21

If \(f\) is continuous on \([-a, a]\), show that \(\int_{-a}^{a} f\left(x^{2}\right) d x=2 \int_{0}^{a} f\left(x^{2}\right) d x\).

Short Answer

Expert verified
\(\int_{-a}^{a} f(x^{2}) dx = 2 \int_{0}^{a} f(x^{2}) dx\).

Step by step solution

01

Separate the integral

First, let's separate the given integral into two integrals over the integration intervals \([-a, 0]\) and \([0, a]\). So \(\int_{-a}^{a} f(x^{2}) dx = \int_{-a}^{0} f(x^{2}) dx + \int_{0}^{a} f(x^{2}) dx\).
02

Substitution method

For the first integral in step 1 (over \([-a, 0]\)), we can carry out a substitution. Let's set \(x=-t\) or \(t=-x\) (where \(x\) is in the interval \([-a, 0]\), thus \(t\) is in the interval \([0, a]\)). Applying this substitution rule, our integral becomes \(\int_{0}^{a} f((-t)^{2})dt\). For \(f((-t)^{2})\), as the function \(f\) is squared, it remains the same, so we have \(f(t^{2})\). This gives us \(\int_{0}^{a} f(t^{2}) dt\). Note that this is the same as the second integral obtained in step 1 (over \([0, a]\)).
03

Final conclusion

Putting together both integrals obtained in steps 1 and 2, we have \(\int_{-a}^{a} f(x^{2}) dx = \int_{0}^{a} f(t^{2}) dt + \int_{0}^{a} f(x^{2}) dx\). This can be simplified as \(2 \int_{0}^{a} f(x^{2}) dx\). Hence, we have proven that \(\int_{-a}^{a} f(x^{2}) dx = 2 \int_{0}^{a} f(x^{2}) dx\).

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