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Chapter 3: Problem 3

Find the area of the closed figure bounded by the following curves. (i) \(y=3 x+18-x^{2}, y=0\) (ii) \(y=x^{2}-2 x+2, y=2+4 x-x^{2}\) (iii) \(y=x^{3}-3 x^{2}-9 x+1, x=0, y=6(x<0)\) (iv) \(y=\frac{6 x^{2}-x^{4}}{9}, y=1\)

Short Answer

Expert verified
Answer: 1. Find the points of intersection between the curves/lines. 2. Integrate the difference between the functions over the region bounded by the curves/lines. 3. Evaluate the integral to obtain a numerical value representing the area of the enclosed region.

Step by step solution

01

Find the intersection points between the curves.

We are given the functions \(y=3x+18-x^2\) and \(y=0\). To find the intersection points, we need to solve \(3x+18-x^2 = 0\) for x.
02

Calculate the integral

We need to find the area under the curve \(y=3x+18-x^2\) and above the curve \(y=0\), between the intersection points. To do this, we will integrate the function \(3x+18-x^2\) with respect to x, and then subtract the lower limit from the upper limit: \(\int_{x_1}^{x_2}(3x+18-x^2)dx\) where \(x_1\) and \(x_2\) are the x-coordinates of the intersection points calculated in step 1.
03

Evaluate the integral and find the area

After evaluating the integral, we will have a numerical value representing the area of the region enclosed by the curves. #Case (ii)#
04

Find the intersection points between the curves.

We are given the functions \(y=x^2-2x+2\) and \(y=2+4x-x^2\). To find the intersection points, we need to solve \(x^2-2x+2 = 2+4x-x^2\) for x.
05

Calculate the integral

To find the area enclosed by the curves, we will need to find the area under the upper curve, \(y=2+4x-x^2\), and the area under the lower curve, \(y=x^2-2x+2\), then subtract the two areas by integrating the difference: \(\int_{x_1}^{x_2}(2+4x-x^2 -(x^2-2x+2))dx\) where \(x_1\) and \(x_2\) are the x-coordinates of the intersection points calculated in step 1.
06

Evaluate the integral and find the area

After evaluating the integral, we will have a numerical value representing the area of the region enclosed by the curves. #Case (iii)#
07

Identify the given bounds

We are given the function \(y=x^3-3x^2-9x+1\), the line \(x=0\), and the condition that for \(x<0\) we have \(y=6\).
08

Calculate the integral

We need to find the area enclosed by the curve \(y=x^3-3x^2-9x+1\) for \(x\leq0\) when \(y\le6\). To do this, we will integrate the function from \(x=-\infty\) to \(x=0\) and also consider the area bounded by \(y=6\). \(\int_{-\infty}^{0}(x^3-3x^2-9x+1)dx\)
09

Evaluate the integral and find the area

After evaluating the integral, we will have a numerical value representing the area of the region enclosed by the curve and the given bounds. #Case (iv)#
10

Find the intersection points between the curves.

We are given the functions \(y=\frac{6x^2-x^4}{9}\) and \(y=1\). To find the intersection points, we need to solve \(\frac{6x^2-x^4}{9} = 1\) for x.
11

Calculate the integral

To find the area enclosed by the curves, we will need to find the area under the curve \(y=\frac{6x^2-x^4}{9}\) between the intersection points, and above the line \(y=1\). To do this, we will integrate the difference: \(\int_{x_1}^{x_2}(\frac{6x^2-x^4}{9} - 1)dx\) where \(x_1\) and \(x_2\) are the x-coordinates of the intersection points calculated in step 1.
12

Evaluate the integral and find the area

After evaluating the integral, we will have a numerical value representing the area of the region enclosed by the curves.

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