You cannot just integrate on [0,2] to get the area. That ignores the part between the curve and the line on the interval [2,3]. That is
∫[2,3] x^3-(27x-54) dx = 11/4
Add that to your area of 4, and you get 27/4
Point P in the curve y=x^3 has coordinates (3,27) and PQ is the tangent to the curve at P.Point Q touches the x-axis.Find the area of the region enclosed between the curve, PQ and the x-axis.
My answer:
I used differenttiation to find the gradient of the tangent: dy/dx=3x^2.
when x=3, the gradient is 3(3)^2=27
Then I used y=mx+c to find the equation of PQ,so 27=27(3)+c and got that the equation is y=27x-54. Finally, I put y=0 in the equation to find x at point Q and got x=2. I intergrated y=x^3 using the limits 2 and 0 and got 4 units^2 for the area. But the answer should be 27/4???? Could you please help me? Thank you
2 answers
The area could also be considered using horizontal strips. In that case, each strip lies between the line and the curve, with no break in the boundary. Then the area is
∫[0,27] (y/27 + 2)-∛y dy = 27/4
∫[0,27] (y/27 + 2)-∛y dy = 27/4
2 answers