First slide
Methods of integration
Question

Let P(x) be a polynomial of least degree whose graph has three points of inflextion (–1, –1), (1,1) and a point with abscissa 0 at which the curve is inclined to the axis of abscissas at an angle of 60

Difficult
Solution

we have P′′(x)=a(x+1)x(x1)

=ax3xP(x)=a14x412x2+bP(x)=a120x516x3+bx+c

As

1=a120+16b+c1=a12016+b+cc=0

and 1=760a+b

Also 3=b

a=60(31)/7

we have 

01P(x)dx=130a+12b

=27(13)+32

 

 

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