Question

Suppose \(f\) is an even function with \(\int_{0}^{8} f(x) d x=9\) Evaluate each integral. a. \(\int_{-1}^{1} x f\left(x^{2}\right) d x .\) b. \(\int_{-2}^{2} x^{2} f\left(x^{3}\right) d x\)

Short answer

Question: Given that $f$ is an even function and $\int_0^8 f(x)dx = 9$, evaluate the following integrals: a) $\int_{-1}^{1} x f\left(x^{2}\right) d x$ b) $\int_{-2}^{2} x^{2} f\left(x^{3}\right) d x$ Answer: a) $\int_{-1}^{1} x f\left(x^{2}\right) d x=0$ b) $\int_{-2}^{2} x^{2} f\left(x^{3}\right) d x=0$

Step-by-step solution

Evaluate Integral a

For the first integral, let's use substitution. Let \(u = x^2\) so that \(du = 2x \, dx\). Then, the integral can be written as: $$ \int_{-1}^{1} x f(u) \frac{1}{2} du . $$ Since \(f(u)\) is even, we have, $$ \int_{-1}^{1} x f(u) \frac{1}{2} du = \frac{1}{2}\int_{-1}^{1} x f(u) du = \frac{1}{2}\left(\int_{-1}^0 (1) x f(u) du + \int_0^1 (-1) (-x) f(u) du\right) = \frac{1}{2}\left(\int_{-1}^0 x f(u) du - \int_0^1 x f(u) du\right) = 0 $$ This is true because the integrals on the intervals \([-1,0]\) and \([0,1]\) are equal and opposite, due to the function being even.

Evaluate Integral b

For the second integral, let's use substitution again. Let \(v = x^3\) so that \(dv = 3x^2 \, dx\). Then, the integral can be rewritten as: $$ \int_{-2}^{2} x^2 f(v) \frac{1}{3} dv . $$ Since \(f(v)\) is even, we have $$ \int_{-2}^{2} x^2 f(v) \frac{1}{3} dv = \frac{1}{3}\int_{-2}^{2} x^2 f(v) dv = \frac{1}{3}\left(\int_{-2}^0 x^2 f(v) dv + \int_0^2 x^2 f(v) dv\right) = $$ $$ \frac{1}{3}\left(\int_{-2}^0 (-x)^2 f(v) dv + \int_0^2 x^2 f(v) dv\right) = \frac{1}{3}\left(\int_{-2}^0 x^2 f(v) dv + \int_0^2 x^2 f(v) dv\right) = 0 $$ Again, this is true because the integrals on the intervals \([-2,0]\) and \([0,2]\) are equal, since we are integrating an even function. So, the final integral values are: a) \(\int_{-1}^{1} x f\left(x^{2}\right) d x=0\) b) \(\int_{-2}^{2} x^{2} f\left(x^{3}\right) d x=0\)

Key concepts

Integration by Substitution

Integration by substitution is a method used to simplify complex integral problems, making them easier to solve. The idea is to substitute a part of the integral with a new variable. This changes the complexity into a simpler form.
In the first integral, we apply substitution by setting \( u = x^2 \) and therefore, the differential term becomes \( du = 2x \ dx \). This substitution turns the integral into \( \int x f(u) \frac{1}{2} du \), simplifying the integration process.
For the second integral, we set \( v = x^3 \), giving us \( dv = 3x^2 \ dx \). This transforms the expression into \( \int x^2 f(v) \frac{1}{3} dv \).
This substitution not only simplifies the integral but also helps to handle the algebra and ensure the integration is correctly executed. The whole point of this technique is to reorganize the function into something we can more easily integrate.

Definite Integral

A definite integral calculates the net area under the curve of a function within defined limits. It provides a numerical value that represents this area.
In the given function \( f(x) \), a definite integral like \( \int_{a}^{b} f(x) \ dx \) represents the area from \( x = a \) to \( x = b \). For an even function \( f \), this area can simplify computations because symmetry reduces the need to calculate both sides separately.
For tasks like \( \int_{-1}^{1} x f(x^2) \ dx \) and \( \int_{-2}^{2} x^2 f(x^3) \ dx \), the symmetry of the function means the areas over negative and positive intervals can cancel out. This is why understanding the nature of a definite integral is crucial; it points us towards recognizing these cancellations and simplifying the integration results.

Symmetry in Integrals

Symmetry in integrals is a powerful tool that leverages the properties of functions to simplify the calculation of integrals. If a function is even, like \( f(x) = f(-x) \), it means the function is symmetric about the y-axis.
When you have an even function and you are calculating a definite integral over a symmetric interval, like \([-a, a]\), this symmetry can result in simple integral evaluations.

• For example, \( \int_{-a}^{a} f(x) \ dx = 2 \int_{0}^{a} f(x) \ dx \) for even functions.
• In our integrals like \( \int_{-1}^{1} x f(x^2) \ dx \), symmetry ensures the negative and positive parts of the interval cancel out the integral to zero.

This understanding can save a lot of time and energy during calculations, making integration more intuitive when these conditions apply.