y=4x2-x3 what is the area between the x axis & the curve??

Answer 1

im assuming those numbers after the x are exponents....

set y=0 solve for x, which is 0 and 4, then integrate y, which is 4/3x^3-1/4x^4, then substitue 4 into that equation, then 0 into that equation and subtract them. so its [4/3(4)^3-1/4(4)^4] - [4/3(0)^3-1/4(0)^4]

Answer 2

First, we need to find where the curve intersects with x-axis. x³ - 5x² + 4x = 0 x (x² - 5x + 4) = 0 x (x-1) (x-4) = 0 x = 0, 1, or 4 Between 0 and 1, y > 0 Between 1 and 4, y < 0 To find area we need to integrate y from 0 to 1 and integrate -y from 1 to 4 A = ∫₀¹ (x³ - 5x² + 4x) dx + ∫₁⁴ (-x³ + 5x² - 4x) dx A = (x⁴/4 - 5x³/3 + 2x²) |₀¹ + (-x⁴/4 + 5x³/3 - 2x²) |₁⁴ A = {(1/4 - 5/3 + 2) - 0} + {(-64 + 320/3 -32) - (-1/4 + 5/3 - 2)} A = 1/4 - 5/3 + 2 - 96 + 320/3 + 1/4 - 5/3 + 2 A = 310/3 - 92 + 1/2 A = 71/6 A = 11.8333...

Answer 3

y = 4x² - x³ = x²(4 - x)

The zeros are 0 and 4 and the function is positive in between. So integrate from 0 to 4.

∫(4x² - x³)dx = 4x³/3 - x^4/4 [eval from 0 to 4]
= 4*64/3 - 256/4 - 0 = 256/3 - 256/4 = 256/12 = 64/3 = 21 1/3

Answer 4

thats for you to figure out since assuming your teacher assigned you (your class) the question - not us - could be 2, as in 2 more points haha