Find area under the curve? Pl refer to the question in the img attached

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Find area under the curve? Pl refer to the question in the img attached

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Answer 1 · 2017-12-24T19:20:49

$A = 49$ $A=49$

Explanation:

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This is a piece-wise function. We have to find the area under each piece separately and add them up.

$y = - 1$ $y=-1$

The area under this piece from $- 2$ $-2$ to $0$ $0$ is:

$A_{1} = \int_{- 2}^{0} - d x = - x = - (0) - (- (- 2)) = 0 - 2 = - 2$ $A_1=int_-2^0 -dx=-x=-(0)-(-(-2))=0-2=-2$

$y = {(x + 2)}^{2}$ $y=(x+2)^2$

The area under this piece from $0$ $0$ to $3$ $3$ is:

$A_{2} = \int_{0}^{3} {(x + 2)}^{2} d x$ $A_2=int_0^3(x+2)^2dx$

Let $u = x + 2$ $u=x+2$ , then $d u = d x$ $du=dx$

$A_{2} = \int u^{2} d u = \frac{1}{3} u^{3} = \frac{1}{3} {(x + 2)}^{3} = \frac{1}{3} {(3 + 2)}^{3} - \frac{1}{3} {(0 + 2)}^{3}$ $A_2=intu^2du=1/3u^3=1/3(x+2)^3=1/3(3+2)^3-1/3(0+2)^3$

$A_{2} = \frac{125}{3} - \frac{8}{3} = \frac{117}{3} = 39$ $A_2=125/3-8/3=117/3=39$

$y = x$ $y=x$

The area under the curve for this piece from $3$ $3$ to $5$ $5$ is:

$A_{3} = \int_{3}^{5} x d x = \frac{1}{2} x^{2} = \frac{1}{2} {(5)}^{2} - \frac{1}{2} {(3)}^{2} = \frac{25}{2} - \frac{9}{2} = \frac{16}{2} = 8$ $A_3=int_3^5xdx=1/2x^2=1/2(5)^2-1/2(3)^2=25/2-9/2=16/2=8$

Area under the piece-wise function is:

$A = A_{1} + A_{2} + A_{3} = 2 + 39 + 8 = 49$ $A=A_1+A_2+A_3=2+39+8=49$

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Find area under the curve? Pl refer to the question in the img attached

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