Another integration problem

Hi guys, I am struggling to solve 1) G in this MAT, below is a link (its a legit safe link directly from maths dept. of Oxford University). I have absolutely no clue how is that sketched graph related to the integral. Any ideas how to solve it?

https://www.maths.ox.ac.uk/system/files/attachments/MAT2011.pdf

It is a trick question!!
From the graph, we know
f(x) =0, x > 1
f(x) = - x + 1, 0 <= x<= 1
f(x) = x +1, -1<= x<0
f(x) = 0, x< -1

The question is find integral f(x^2 -1)dx, x from -1 to 1.
The trick is to find f(x^2 -1).
We will do this integral in two parts.
1. x from 0 to 1
At this interval, x^2 -1 will be in [-1, 0],
thus this part of the graph will be used: y = x + 1
Thus, f(x^2 - 1) = x^2 -1 + 1 = x^2
Int f(x^2 - 1) dx from 0 to 1
= Int x^2 dx, from 0 to 1
= 1/3 x^3, from 0 to 1
= 1/3

2. x from -1 to 0
At this interval, x^2 -1 will also be in [-1, 0],
thus this same part of the graph will be used: y = x + 1
Thus, f(x^2 - 1) = x^2 -1 + 1 = x^2
Int f(x^2 - 1) dx from 0 to 1
= Int x^2 dx, from 0 to 1
= 1/3 x^3, from 0 to 1
= 1/3 ------- same answer

Hence, the integral = 1/3 + 1/3 = 2/3

The graph for y=f(x)? It appears as a piece-wise function. There are four "pieces". Look at this piece when 0≤x≤1. It is a straight line with slope of -1. Thus, y = f(x) =-x+1. Observe that these two points: (0,1) and (1,0) are "end points" which satisfies y = -x+1. Let me know if this helps.

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