There is a diagram with the curve y=(2x-5)^4.The point P has coordinates (4,81) and the tangent to the curve at P meets the x-axis at Q.Find the area of the region enclosed between curve ,PQ and the x-axis.

Answer 1

y = (2x-5)^4
dy/dx = 4(2x-5)^3 (2) = 8(2x-5)^3
P(4,81) lies on the curve, and the slope of the tangent at that point is
8(3^3) = 216
and the equation of the tangent is
y-81 = 216(x-4)
y = 216x - 783
At Q, the intercept with the x-axis, x = 783/216 = 29/8

the curve makes contact with the x-axis at 5/2

Let a vertical from P cut the x-axis at R

So find the area of the shape between the x-axis the vertical PR and the curve, then subtract the area of triangle PRQ.

you will need the integral of (2x-5)^4 which would be (1/10)(2x-5)^5

take over

Answer 2

How come At Q, the intercept with the x-axis, x = 783/216 = 29/8

Answer 3

How does the vertical cut x-axis?

Answer 4

From the tangent equation, y = 216x - 783
at the x-axis, y = 0
so 0 = 216x - 783
216x = 783 ----> x = 783/216 = 29/8

your 2nd post: the x-axis is horizontal, any vertical line would cut it, wouldn't it?

Answer 5

I found area of triangle but couldn't get area of the shape between the x-axis the vertical PR and the curve

Answer 6

I got it.Thanks