if anyone uses Pure mathematics 1 by Hugh Neill & Douglas Quadling can u plz turn over to page 253 Ex.16D Q4.
i got rong answer prolly cause one of my limit is rong and thts whr im confused, can summon plz do n lemme kno?
if u still want the answer, here it is (i dont know how to use the integral sign, sorry)
ive attached a drawing, (please refer to that while reading the explanation)
first of all we'll find the area K+L by integration, we then subtract the area L from this area to find the area K.
first of all we find the area K+L by integration----->>>
the integral of the function (2x-5)^4---> [(2x-5)^5)/10]
the limits will be b/w Q and R, we can find the co-ordinate of Q by equating the equation of the curve to 0.
then the area K+L comes out to be---->
=> 24.3-0=24.3
we then find the area L. (we first have to find the equation of PR though).
since f(x)=(2x-5)^4
f'(x)=[4(2x-5)^3]*2
therefore the gradient of the tangent=216 at (4,81)
then the equation of PQ becomes---> y=216x-783, we find out the x-coordinate of R by equating the equation of PQ to 0.
then we find the area L-->
0.5*0.375*81=15.1875
therefore K=24.3-15.1875=9.1125
hope i helped
p.s. sorry about the bad handwriting in the drawing, so just for reference Q(2.5,0), P(4,81), R(4,0), S(3.625,0)