The corresponding areas under the curves for y = 2x-x² in the interval [1, 2] and y = x³-6x²+8x in the interval [0, 4] are 2/3 unit² and 0 unit²
Given:
1. y = 2x-x² in the interval [1, 2]
2. y = x³-6x²+8x in the interval [0, 4]
In order to find the area under the curve for these functions for the given intervals, we will take the integral of the functions and use the intervals as upper and lower limits: Thus:
y = 2x-x² in the interval [1, 2] will be:
[tex]\int\limits^2_1 {2x-x^{2} } \, dx = \left[\frac{ 2x^2}{2}-\frac{ x^3}{3}\right]_1^2[/tex]
[tex]= \left[x^{2} -\frac{ x^3}{3}\right]_1^2[/tex]
Substitute the value of the upper and lower limits into x:
= [2²- 2³/3] - [1²- 1³/3]
= [4 - 8/3] - [ 1 - 1/3]
= [4/3] - [2/3]
= 2/3 unit²
y = x³-6x²+8x in the interval [0, 4] will be:
[tex]\int\limits^4_0 {x^{3} -6x^{2} + 8x } \, dx = \left[\frac{ x^4}{4}-\frac{ 6x^3}{3}+\frac{ 8x^2}{2}\right]_0^4[/tex]
[tex]= \left[\frac{ x^4}{4}- 2x^3+4x^2\right]_0^4[/tex]
= [4⁴/4 - 2(4)³ + 4(4)²] - [0⁴/4 - 2(0)³ + 4(0)²]
= [ 64 - 128 + 64] - [0 - 0 -0]
= 0 unit²
Therefore, the areas under the curves y = 2x-x² in the interval [1, 2]
and y = x³-6x²+8x in the interval [0, 4] are 2/3 unit² and 0 unit² respectively
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