query 32

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course Mth 272

8/8 12:29am

032.

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Question: `qQuery problem 7.8.6 integrate (x^2+y^2) with respect to y from x^2 to `sqrt(x)

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Your solution:

An antiderivative would be x^2 y + y^3 / 3.

Evaluating this antiderivative at endpoints y = x^2 and y = `sqrt(x) we get

[ x^2 * `sqrt(x) + (`sqrt(x) ) ^ 3 / 3 ] - [ x^2 * x^2 + (x^2)^3 / 3 ] =

x^(5/2) + x^(3/2) / 3 - ( x^4 + x^6 / 3) =

x^(5/2) + x^(3/2) / 3 - x^4 - x^6 / 3.

This can be simplified in various ways, but the most standard form is just decreasing powers of x:

- x^6/3 - x^4 + x^(5/2) + x^(3/2)/3.

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Given Solution:

`a An antiderivative would be x^2 y + y^3 / 3.

Evaluating this antiderivative at endpoints y = x^2 and y = `sqrt(x) we get

[ x^2 * `sqrt(x) + (`sqrt(x) ) ^ 3 / 3 ] - [ x^2 * x^2 + (x^2)^3 / 3 ] =

x^(5/2) + x^(3/2) / 3 - ( x^4 + x^6 / 3) =

x^(5/2) + x^(3/2) / 3 - x^4 - x^6 / 3.

This can be simplified in various ways, but the most standard form is just decreasing powers of x:

- x^6/3 - x^4 + x^(5/2) + x^(3/2)/3.

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Question: `qQuery problem 7.8.14 integrate `sqrt(1-x^2) from 0 to x wrt y then from 0 to 1 wrt x

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Your solution:

The limits on the first integral are 0 and x.

The result of the first integral is then to be integrated with respect to x.

The solution:

** Since the function has no y dependence the whole expression is treated as a constant and an antiderivative with respect to y is just `sqrt(1 - x^2) * y. The definite integral from 0 to x is therefore `sqrt(1 - x^2) * x - `sqrt(1 - x^2) * 0 = x sqrt(1-x^2).

We now need to integrate the resulting expression x sqrt(1 - x^2) with respect to x, from x = 0 to x = 1.

An antiderivative of x sqrt( 1 - x^2 ) is easily found by letting u = 1 - x^2, so that du = -2 x dx and x dx = -du / 2.

The integrand becomes -sqrt(u), with antiderivative -2/3 u^(3/2). Substituting 1 - x^2 for u we get antiderivative -2/3 (1 - x^2)^(3/2). Between x = 0 and x = 1 this expression changes from -2/3 to 0, a change of 2/3.

The definite integral is therefore 2/3.

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Given Solution:

`a The limits on the first integral are 0 and x.

The result of the first integral is then to be integrated with respect to x.

The solution:

We now need to integrate the resulting expression x sqrt(1 - x^2) with respect to x, from x = 0 to x = 1.

An antiderivative of x sqrt( 1 - x^2 ) is easily found by letting u = 1 - x^2, so that du = -2 x dx and x dx = -du / 2.

The definite integral is therefore 2/3.

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Question: `qQuery problem 7.8.32 sketch region and reverse order of integration for integral of 1 from 0 to 4-y^2, then from -2 to 2.

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Your solution:

We first integrate 1 from 0 to 4 - y^2, with respect to x. An antiderivative of 1 with respect to x is x; substituting the limits 0 and 4 - y^2 we get [ 4 - y^2 ] - 0, or just 4 - y^2.

Then integrating from y = -2 to 2, with respect to y, the antiderivative of 4 - y^2 is 4 y - y^3 / 3; substituting the limits -2 and 2 we get [ 4 * 2 - 2^3 / 3 ] - [ 4 * (-2) - (-2)^3 / 3 ] = 8 - 8/3 - [ -8 - (-8 / 3) ] = 16 - 16/3 = 32 /3.

The region of integration is -2 <= y <= 2, 0 <= x <= 4 - y^2. This region lies between the parabola x = 4 - y^2 and the y axis. The vertex is (4, 0), the parabola opens to the left, and it intercepts the y axis at the points (0, 2) and (0, -2).

For a given x value between 0 and 4, a vertical line through (x, 0) intersects this region at the points where x = 4 - y^2. The y coordinates of these points are easily found by solving x = 4 - y^2 for y, obtaining y = +-sqrt(4 - x). Thus the vertical line intersects the region between (x, -sqrt(4-x)) and (x, sqrt(4 - x)). If x < 0, or if x > 4 the point (x, 0) lies outside the region.

Thus the region can be described by

0 <= x <= 4, -sqrt(4-x) <= y <= sqrt(4 - x)

and we integrate 1 first with respect to y, from limit -sqrt(4 - x) to sqrt( 4 - x), then from x = 0 to 4.

The 'inner integral' with respect to y yields antiderivative y, which evaluated between the limits -sqrt(4 - x) and sqrt( 4 - x) gives us 2 sqrt(4 - x).

This result is then integrated between x = 0 and x = 4. Our antiderivative is easily found to be -2 * (2/3 (4 - x)^(3/2)) = -4/3 (4 - x)^(3/2). Evaluating between the limits x = 0 and x = 4 we obtain -4/3 (4 - 4)^(3/2) - (-4/3 * (4 - 0)^(3/2)) = 32/3.

This result agrees with our first integral, as expected.

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Given Solution:

`a We first integrate 1 from 0 to 4 - y^2, with respect to x. An antiderivative of 1 with respect to x is x; substituting the limits 0 and 4 - y^2 we get [ 4 - y^2 ] - 0, or just 4 - y^2.

Thus the region can be described by

0 <= x <= 4, -sqrt(4-x) <= y <= sqrt(4 - x)

and we integrate 1 first with respect to y, from limit -sqrt(4 - x) to sqrt( 4 - x), then from x = 0 to 4.

The 'inner integral' with respect to y yields antiderivative y, which evaluated between the limits -sqrt(4 - x) and sqrt( 4 - x) gives us 2 sqrt(4 - x).

This result agrees with our first integral, as expected.

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Question: `qQuery problem 7.8.40 Area beneath curve 1/`sqrt(x-1) for 2 <= x <= 5

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Your solution:

The region goes from x = 2 to x = 5, and from the x axis to the curve y = 1 / `sqrt(x-1).

To find the area you integrate 1 over the region.

The integral would be from y = 0 to y = 1/`sqrt(x-1) with respect to y (this is the inside integral), then from x = 2 to x = 5 with respect to x (the outside integral).

Inside integral: antiderivative of 1 is y; substituting limits we have [1/`sqrt(x-1)] - 0 = 1 / `sqrt(x-1).

Outside integral is now of 1/`sqrt(x-1) with respect to x, from x = 2 to x = 5. Antiderivative is 2 `sqrt(x-1) which can be found thru the u substitution u = x - 1. Evaluating at limits we have 2 `sqrt(5-1) - 2 `sqrt(2-1) = 2 `sqrt(4) - 2 `sqrt(1) = 2.

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Given Solution:

`a The region goes from x = 2 to x = 5, and from the x axis to the curve y = 1 / `sqrt(x-1).

To find the area you integrate 1 over the region.

The integral would be from y = 0 to y = 1/`sqrt(x-1) with respect to y (this is the inside integral), then from x = 2 to x = 5 with respect to x (the outside integral).

Inside integral: antiderivative of 1 is y; substituting limits we have [1/`sqrt(x-1)] - 0 = 1 / `sqrt(x-1).

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Question: `qQuery problem 7.8.44 area bounded by xy=9, y=x, y=0, x=9

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Your solution:

xy = 9 means y = 9/x; graph decreases at decreasing rate from asymptote at y axis toward asymptote at x axis.

y = x is straight line at 45 deg to x axis.

y = x intersects xy = 9 when x^2 = 9 or x = 3 (x^2 = 9 found by substituting y = x into xy = 9).

The graph is bounded by y = 0, which is the x axis. Also by x = 9. So the region lies above the x axis, below the y = x curve from x = 0 to x = 3 ( where y = x is lower than y = 9 / x) and below the xy = 9 curve from x = 3 to x = 9.

The description:

0 <= x <= 3, 0 <= y <= x

plus

3 <= x <= 9, 0 <= y <= 9/x.

We are integrating area so for our first region:

Our 'inner' integral is the integral of 1 with respect to y, between the limits y = 0 and y = x. Our antiderivative is therefore y, which changes between the limits from 0 to x, a change of x. So our 'inner' integral gives us x.

Integrating this result x with respect to x, between limits x = 0 and x = 3, our antiderivative is x^2 / 2. Between the limits x = 0 and x = 3 our antiderivative changes from 0^2 / 2 = 0 to 3^2 / 2 = 9/2, so the change in the antiderivative is 9/2.

For our second region:

Our 'inner' integral is the integral of 1 with respect to y, between the limits y = 0 and y = 9 / x. Our antiderivative is y, which changes between the limits from 0 to 9 / x, a change of 9/x. So our 'inner' integral gives us 9 / x.

Integrating this result 9 / x with respect to x, between limits x = 0 and x = 3, our antiderivative is 9 ln | x |. Between the limits x = 3 and x = 9 our antiderivative changes from 9 ln ( 3) to 9 ln ( 9 ), a change of 9 ln(9) - 9 ln(3) = 9 ln(9/3) = 9 ln(3).

The area of our region is therefore 9/2 + 9 ln(3), which we simplify to 9 ( ln(3) + 1/2 ).

So we first integrate 1 from x = 0 to x = 3 (outer integral) and from y = 0 to y = x (inner integral). Inner integral gives antiderivative y; substituting limits we get x - 0 = x. Integrating x from x = 0 to x = 3 we get antiderivative x^2 / 2; substituting limits yields 3^2 / 2 - 0^2 / 2 = 9/2 (which is the area of the triangle formed between x = 0 and x = 3).

Then we integrate 1 from x = 3 to x = 9 (outer integral) and from y = 0 to y = 9 / x (inner integral). Inner integral gives antiderivative y; substituting limits we get 9/x - 0 = 9/x. Integrating x from x = 3 to x = 9 we get antiderivative 9 ln | x | ; substituting limits yields 9 ln | 9 | - 9 ln | 3 | = 9 ln | 9/3 | = 9 ln | 3 |.

Total area is therefore 9 ln | 3 | + 9/2 = 9 ( ln | 3 | + 1/2 ).

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Given Solution:

xy = 9 means y = 9/x; graph decreases at decreasing rate from asymptote at y axis toward asymptote at x axis.

y = x is straight line at 45 deg to x axis.

y = x intersects xy = 9 when x^2 = 9 or x = 3 (x^2 = 9 found by substituting y = x into xy = 9).

The description:

0 <= x <= 3, 0 <= y <= x

plus

3 <= x <= 9, 0 <= y <= 9/x.

We are integrating area so for our first region:

For our second region:

The area of our region is therefore 9/2 + 9 ln(3), which we simplify to 9 ( ln(3) + 1/2 ).

Total area is therefore 9 ln | 3 | + 9/2 = 9 ( ln | 3 | + 1/2 ).

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Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment.

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