How do you find the area bounded by the x axis, y axis, $y = x^{2} + 1$ $y=x^2+1$ and $x = 2$ $x=2$ ?

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How do you find the area bounded by the x axis, y axis, $y = x^{2} + 1$ $y=x^2+1$ and $x = 2$ $x=2$ ?

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Answer 1 · 2017-09-25T17:13:09

$\frac{14}{3}$ $14/3$

Explanation:

graph{y=x^2+1 [-8.75, 11.25, -1.52, 8.48]}
limit of x is from $x = 0$ $x=0$ to $x = 2$ $x=2$
Function is $y = x^{2} + 1$ $y=x^2+1$

Area Between the curve and X-axis is
$A r e a = \int_{0}^{2} y d x = \int_{0}^{2} (x^{2} + 1) d x$ $Area=int_0^2ydx=int_0^2(x^2+1)dx$

$A r e a = {[\frac{x^{3}}{3} + x]}_{0}^{2} = [\frac{2^{3}}{3} + 2 - 0 - 0] = \frac{8}{3} + 2 = \frac{14}{3}$ $Area=[x^3/3+x]_0^2=[2^3/3+2-0-0]=8/3+2=14/3$

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How do you find the area bounded by the x axis, y axis, $y = x^{2} + 1$ $y=x^2+1$ and $x = 2$ $x=2$ ?

1 Answer

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How do you find the area bounded by the x axis, y axis, y=x^2+1y=x2+1y=x^2+1 and x=2x=2x=2?

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How do you find the area bounded by the x axis, y axis, $y = x^{2} + 1$ $y=x^2+1$ and $x = 2$ $x=2$ ?