Find the area of the region bounded by the parabola y = 4x2, the tangent line to this parabola at (4, 64), and the x-axis.

Question

Find the area of the region bounded by the parabola y = 4x², the tangent line to this parabola at (4, 64), and the x-axis.

Answer 1

Area under y = 4x² = ₀∫⁴ (4x²) dx = (4/3)x³ evaluated between 0 and 4 = (4/3)(4³) - 4/3(0³) = 256/3

The tangent line has a slope found by taking the derivative:

y' = 8x and y'(4) = 8(4) = 32 therefore the equation of the tangent line at (4, 64) using the point-slope form of a line is: y - 64 = 32(x - 4) or y = 32x -64. When y = 0, x = 2.

Area under the tangent line is ₂∫⁴(32x - 64) dx = 16x² - 64x evaluated between 2 and 4 = 16(4²) - 64(4) - (16(2²) - 64(2)) = 256 - 256 - 64 + 128 = 64

Area between the curve and the tangent line = 256/3 - 64 = 256/3 - 196/3 = 64/3

Answer 2

The tangent at (4,64) of y = 4x² has slope 8x =8*4 = 32.

Thus, y =mx+b yields 64 =32(4) +b, so b=-64

Then, 0 =32x -64, so x=2.The tangent line crosses the x axis at x =2.

Thus, the area between the curve and the tangent line is the area under the curve minus the area of the triangle with vertices at (2,0),(4, 0), and (4,64)

∫₀⁴ 4x² dx = 4/3x³ |₀⁴ = 4/3*4³ -0 = 256/3

The area of the triangle is 1/2 * 2 * 64 =64

256/3 - 64 = 64/3