Source: MAT $2011$
Above is the graph of the function $f(x)$. It is required to find $$\int_{-1}^{1}f(x^2-1)dx$$ So the function $f(x)$ can be split into two parts -
- $f(x) = \left(x+1\right)$ for $-1 \le x \le 0$
- $f(x) = \left(-x+1\right)$ for $0 \lt x \le 1$
Their transformations would be the following respectively:
- $f(x^2-1) = x^2$ for $-1 \le x \le 0$
- $f(x^2-1) = -x^2 + 2$ for $0 \lt x \le 1$
So I should be able to integrate both of these functions in their respective domains to get my answer. $$\int_{-1}^{1}{f(x^2-1)dx} = \int_{-1}^{0}{(x^2)dx} + \int_{0}^{1}{(-x^2+2)dx} = 2$$
Yeah, my method is completely wrong. How should I do it, and why is my method wrong?
I think your method fails, because you don't change the domain of your piece-wise defined function. Take, for example, $x=0.5$. Therefore, $x^2-1 = -0.75$, meaning: A point thay used to lie on the right side is now on the left side. The domains (especially the borders of those) must be altered to suit this transformation. E.g.: The first part of your domain should actually be $-1 \leq x^2-1 \leq 0$. Do you see the problem with your integral?
I would figure out the new domains for the two parts of the function. For the first part that would be $-1 \leq x^2 -1 \leq 0$, therefore $0 \leq x^2 \leq 1$ and for the second part $0 \lt x^2 -1 \leq 1$, therefore $1 \lt x^2 \leq 2$. Meaning: you have now new inequalities defining your domains. Figure out, for which x each of the inequalities is fulfilled (that is: solve the inequalities for $x$) and you obtain the new final domains for your tranformed function (hint: It will be more than two pieces). Finally, you can integrate between -1 and 1 knowing which function will reside where.