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3 years ago

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Can someone help me with the attached problem it’s for geometry, it’s urgent!

1 answer:

Sedaia [141]3 years ago

7 0

I am sorry but I cannot see the picture properly, can you add another photo? Sorry for the inconvenience.

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Find the area bounded by the given curves: <br> y=2x−x2,y=2x−4

Andrej [43]

Answer:

$A = [\frac{32}{3}]$

Step-by-step explanation:

Given

$y_1 = 2x - x^2$

$y_2 = 2x - 4$

Required

Determine the area bounded by the curves

First, we need to determine their points of intersection

$2x - x^2 = 2x - 4$

Subtract 2x from both sides

$-x^2 = -4$

Multiply through by -1

$x^2 = 4$

Take square root of both sides

$x = 2$ or $x = -2$

This Area is then calculated as thus

$A = \int\limits^a_b {[y_1 - y_2]} \, dx$

<em>Where a = 2 and b = -2</em>

Substitute values for $y_1$ and $y_2$

$A = \int\limits^a_b {(2x - x^2) - (2x - 4)} \, dx$

Open Brackets

$A = \int\limits^a_b {2x - x^2 - 2x + 4} \, dx$

Collect Like Terms

$A = \int\limits^a_b {2x - 2x- x^2 + 4} \, dx$

$A = \int\limits^a_b {- x^2 + 4} \, dx$

Integrate

$A = [-\frac{x^{3}}{3} +4x](2,-2)$

$A = [-\frac{2^{3}}{3} +4(2)] - [-\frac{-2^{3}}{3} +4(-2)]$

$A = [-\frac{8}{3} +8] - [-\frac{-8}{3} -8]$

$A = [\frac{-8+ 24}{3}] - [\frac{8}{3} -8]$

$A = [\frac{-8+ 24}{3}] - [\frac{8-24}{3}]$

$A = [\frac{16}{3}] - [\frac{-16}{3}]$

$A = [\frac{16}{3}] + [\frac{16}{3}]$

$A = [\frac{16 + 16}{3}]$

$A = [\frac{32}{3}]$

Hence, the Area is:

$A = [\frac{32}{3}]$

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stiks02 [169]

X/5 + 7 = 14
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