Question

4
(i) the diagram below shows the graph of ( y = f(x) ) for ( x leq 0 ). it is given that ( f(x) ) is an odd
function such that ( f(-x)=-f(x) ) and the graph cuts the ( x )-axis at ( x = -7, x = -2.5 ) and
( x = 0 ).
(a) sketch the graph of ( y = f(x) ) for ( x>0 ). label the axial intercepts clearly. 1
(b) given further that ( int_{-2.5}^{-7} f(x) d x=-p ), and the area bounded by ( y = f(x) ) and ( x )-axis
from ( x=-7 ) to ( x = 0 ) is ( q ) units ( ^{2} ), where ( p ) and ( q ) are positive constants. find
( int_{0}^{2.5} k f(-x) d x ) in terms of ( k, p ) and/or ( q ). 2

Explanation:

Part (a)

Since \(y = f(x)\) is an odd function, its graph is symmetric about the origin. For \(x>0\), if \((x,y)\) is on the graph of \(y = f(x)\) for \(x < 0\), then \((-x,-y)\) is on the graph for \(x>0\). The \(x -\)intercepts for \(x>0\) are \(x = 2.5\) and \(x=7\) (because if \(f(a)=0\) for \(a<0\), then \(f(-a)=-f(a) = 0\) for \(a>0\)).

Step1: Use the property of odd - function and substitution

Let \(u=-x\), then \(du=-dx\). When \(x = 0\), \(u = 0\); when \(x = 2.5\), \(u=-2.5\). So \(\int_{0}^{2.5}kf(-x)dx=-k\int_{0}^{-2.5}f(u)du=k\int_{-2.5}^{0}f(u)du\).

Step2: Use the relationship between definite - integrals and areas

We know that \(\int_{-7}^{-2.5}f(x)dx=-p\) and \(\int_{-7}^{0}f(x)dx = q\). By the property of definite - integrals \(\int_{-7}^{0}f(x)dx=\int_{-7}^{-2.5}f(x)dx+\int_{-2.5}^{0}f(x)dx\). Then \(\int_{-2.5}^{0}f(x)dx=q + p\).

Answer:

Sketch the graph of \(y = f(x)\) for \(x>0\) with \(x -\)intercepts at \(x = 2.5\) and \(x = 7\). The graph for \(x>0\) is the reflection of the graph for \(x<0\) about the origin.

Part (b)