66-68. Areas of regions (Use of Tech) Find the area of the fol...

Question

66-68. Areas of regions (Use of Tech) Find the area of the following regions.

66. The region bounded by the curve y = (x - x²)/[(x + 1)(x² + 1)] and the x-axis from x = 0 to x = 1

Accepted Answer

Hello there. Today we're going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Find the area of the region bounded by the curve. Y is equal to x minus 1 divided by parentheses X + 2 multiplied by parenthesis of X + 4. And the X axis from X is equal to 2 to X is equal to 3. Awesome. So it appears for this particular problem we're asked to take all the information that is provided to us, and we're asked to find the area of the region bounded by this curve for this expression of Y. Based on what we know for our curve Y and what we're told that the x axis is from X is equal to 2 to X is equal to 3, so we're ultimately trying to find this area. So we're also given a figure to help us better visualize what's happening in this particular prompt, and we have our basic XY coordinate plane where our Y axis is the vertical axis, and it's from -0.5 to 0.5. And then we have our X-axis, which is the which is our horizontal axis, and we know that our vertical is Y and our horizontal is X, and it goes from our origin. 00 to 4. And as we're told, we have Our curve Y, which is equal to X minus 1 divided by parentheses X + 2 multiplied by parentheses X + 4, and our curve Y is represented in blue, and then we have our shaded region which represents the. Area, so the shaded blue region represents the area of the region bounded by our curve. So that's the area that we're ultimately trying to solve for. And as we can see, our curve is located between the two other curves for X is equal to 2 and X is equal to 3, which are two vertical. Curves, they're two straight lines that are vertical for the curves, and our curve Y is located between the curves X is equal to 2 and X is equal to 3. So now that we know that we're trying to solve for the shaded blue region. Our first step that we need to take is we need to check the sign of Y on the closed interval of 2.3. So whenever you see square brackets, like the square brackets, it means you're dealing with a closed interval. If they're parenthesis, it would be an open interval. So with that in mind, we're checking the sign of Y on this closed interval of 2.3. We also note from the problem itself that the numerator states that X is -1 and the denominator is parentheses X + 2 multiplied by parentheses X + 4. Fantastic. So with that in mind, We need to note that on the closed interval of square brackets 2.3, we can go ahead and state that x minus 1 is greater than 0, cents, which I'll just calls for short, 2 minus 1 is equal to 1, which 1 is greater than 0. So that will mean that x + 2 is greater than 0 and X + 4 is greater than 02. So therefore, Y is greater than 0 on the closed interval of 2.3. So that will mean, as we should recall, the area A will be equal to the integral evaluated from 2 to 3 of YDX, which is equal to the integral evaluated from 2 to 3 of X minus 1. Divided by parentheses X + 2 multiplied by parentheses X + 4. DX fantastic. So moving right along here, our second step now is we need to recall and use partial fraction decomposition, and we can therefore write when we use partial fraction decomposition, we can write that X minus 1 divided by parentheses X + 2 multiplied by parentheses X + 4. will be equal to a divided by X + 2 + B divided by X + 4. So we now need to multiply both sides by the denominator, and when we do, we will find that X minus 1 is equal to a multiplied by parentheses X + 4. Plus B multiplied by parentheses X + 2, which will be equal to A X + 4 A plus B X + 2B. Which will be equal to parentheses A plus B multiplied by X. So A plus B multiplied by X + or A + 2B in parentheses. So now we need to match coefficients. So for the coefficient of X, we have 1 is equal to a plus B. And for the constant term negative one, it will be equal to 4A, so 4 multiplied by A or just 4A. Plus 2 be fantastic. So moving right along here. So, from the first, we will note that B is equal to 1 minus A. So we can substitute this value into the second, so we can find that -1 is equal to 4A plus 2 multiplied by parentheses 1 minus A, which is equal to 4A plus 2 minus 2A, which is equal to 2A plus 2. So, that will mean that 2A is equal to -3, so when we rearrange to isolate and solve for A, you will find that A is simply equal to -3 halves or 3 divided by 2. So, once again, so A is equal to -3 halves or 3 divided by 2. So then we can go ahead and write that B is equal to 1 minus parentheses 3 divided by 2, which is equal to 1 + 3 divided by 2, which is equal to 5 divided by 2 or 5 halves. So now our third step now is we need to write the interval with partial fractions. So when we do that, we will find that the integral evaluated from 2 to 3 of X minus 1 divided by parentheses X + 2 multiplied by parentheses X + 4. DX will be equal to the integral from 2 to 33 divided by 2 multiplied by 1 divided by X plus 2. Plus 5 divided by 2 multiplied by 1 divided by X plus 4. DX Which will be equal to -3 divided by 2 multiplied by the integral from 2 to 3 of 1 divided by X plus 2 DX plus 5 divided by 2 multiplied by the interval from 2 to 3 of 1 divided by X plus 4DX. Fantastic. So our fourth step now is we need to evaluate our integrals. So we need to take the integral from 2 to 3 of 1 divided by X + 2, the X, which will be equal to square brackets of the natural log of the absolute value of X plus 2. Which will be evaluated from 2 to 3. Which will be equal to the natural log of 5 minus the natural log of 4, which will be equal to the natural log of 5 divided by 4. Fantastic. And we will take the integral from 2 to 3 of 1 divided by X plus 4 the X, which will be equal to square brackets of the natural log of the absolute value of X + 4 evaluated from 2 to 3, which will be equal to the natural log of 7. Minus the natural log of 6, which will be equal to the natural log of 7 divided by 6. Fantastic. And our fifth and final step is we need to combine the results, and we will find that the area is equal to -3. Divided by 2, so 3 divided by 2, multiplied by the natural log of 5 divided by 4, plus 5 divided by 2 multiplied by the natural log of 7 divided by 6. And that's it. That is our final answer. Hooray, we did it. So if you do not like having the negative out front, you can rearrange this expression and write the positive value first. You could do 5 divided by 2 multiplied by the natural log of 7 divided by 6 minus 3 divided by 2 multiplied by the natural log of 5 divided by 4. Both answers are correct. So, let's write that down actually. So we could say our final answer written all nice like will be equal to, so A is equal to 5 divided by 2 multiplied by the natural log is 7 divided by 6 minus 3 divided by 2 multiplied by the natural log of 5 divided by 4, and that's it. That is our final answer. Hooray, we did it. So going back to look at the problem itself, our final answer based on everything that we've learned. And to confirm, our final answer without a doubt once again will be. Drum roll. 5 divided by 2 multiplied by the natural log of 7 divided by 6 minus 3 divided by 2 multiplied by the natural log of 5 divided by 4. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

66-68. Areas of regions (Use of Tech) Find the area of the following regions.66. The region bounded by the curve y = (x - x²)/[(x + 1)(x² + 1)] and the x-axis from x = 0 to x = 1

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66-68. Areas of regions (Use of Tech) Find the area of the following regions.
66. The region bounded by the curve y = (x - x²)/[(x + 1)(x² + 1)] and the x-axis from x = 0 to x = 1