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.
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
How come At Q, the intercept with the x-axis, x = 783/216 = 29/8
How does the vertical cut x-axis?
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?
I found area of triangle but couldn't get area of the shape between the x-axis the vertical PR and the curve
I got it.Thanks
To find the area of the region enclosed between the curve, PQ, and the x-axis, we need to integrate the curve from the x-coordinate of point P to the x-coordinate of point Q.
Step 1: Find the x-coordinate of point Q.
Since the tangent to the curve at P meets the x-axis at Q, we know that the y-coordinate of point Q is 0. We can find the x-coordinate of Q by substituting 0 for y in the equation of the curve:
0 = (2x - 5)^4
Solving for x in this equation will give us the x-coordinate of Q.
Step 2: Find the expression for the curve.
Given y = (2x - 5)^4, we can expand it using the binomial theorem to get:
y = 16x^4 - 80x^3 + 150x^2 - 100x + 25
Step 3: Integrate the curve.
To find the area, we need to integrate the curve expression we obtained from step 2 from the x-coordinate of point P to the x-coordinate of point Q. In this case, the interval is from the x-coordinate of (4, 81) to the x-coordinate of point Q.
∫[x-coordinate of P, x-coordinate of Q] (16x^4 - 80x^3 + 150x^2 - 100x + 25) dx
Step 4: Evaluate the integral.
Evaluate the integral obtained from step 3 to find the area enclosed between the curve, PQ, and the x-axis.
Once you evaluate the integral, that will give you the area of the region enclosed between the curve, PQ, and the x-axis.