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This question is to test integration by parts.  First let u=x² and u'=2x as a result, and v'=e^x and so v=e^x too. Then use the by parts formula to express the integral as x...

df/dx = 4x - 2

turning point when differential = 0 

==>

4x = 2 hence;  x = 0.5

When x = 0.5 f(x) is equal to (substitute) 3.5

hence the turning point is at (0.5,3.5)<...

By looking at both equations you spot that the 'x' in the first equation has coefficient equal to 1, so it would be quite convenient to make 'x' the subject of the first equation and then substitute what ...

the curve crosses the graph at the x axis at 0.5,3 and -4 so f(0.5)=0, f(3)=0,f(-4)=0. All linear factors are the form of g(x)=(x-a) so 0=g(0.5)=0.5-a. rearranging we get that a=0.5 ie g(x)=x-0.5 similarl...

Factorise denominator using the factor theorem giving - 1/(x+1)(x+2)(x-1) Use partial fractions to turn this into an easier form to integrate giving - 1/3(x+2) + 1/6(x-1) - 1/2(x+...