How Do You Evaluate an Integral Using Geometric Interpretation?

[TsAmE]

Homework Statement

Evaluate ∫ -2 to 2 (x + 3)(4 - x^2)^1/2 dx by writing it as a sum of 2 integrals and interpreting one of those integrals in terms of an area.

Homework Equations

None.

The Attempt at a Solution

∫ -2 to 2 (x + 3)(4 - x^2)^1/2 dx
= ∫ 0 to -2 (x + 3)(4 - x^2)^1/2 + ∫ 0 to -2 (x + 3)(4 - x^2)^1/2

I then draw a graph which resembled 1/4 of an ellipse (having a major axis on the y-axis), for 0 to 2. The y-intercept = 6 and x intercepts: -3, 2, -2.

However the answer in the back of my book was 6pi, and I don't understand why

 

[thepatient]

Hm... looks like you might have to multiply in the (4-x^2)^1/2 into (x+3), then it would be a simple u-substitution in the first half, and probably a trig sub in the second... Very doable but a little long...

 

[rock.freak667]

$$\int_{-2} ^{2} (x + 3)(4 - x^2)^{\frac{1}{2}} dx$$

Try expanding out (x+3)(4-x²)^1/2, then split the integral.


(Hint: (a+b)c = ac+bc)

 

[Mark44]

Hm... looks like you might have to multiply in the (4-x^2)^1/2 into (x+3), then it would be a simple u-substitution in the first half, and probably a trig sub in the second... Very doable but a little long...

Not long at all if the OP follows the suggestion of interpreting one of the integrals as the area of a geometric object.

 

[thepatient]

Not long at all if the OP follows the suggestion of interpreting one of the integrals as the area of a geometric object.

Oh wow haha, you're right. It just looks like half of a tilted ellipse. It's not so bad then. XD

 

[Mark44]

In the broadest sense, it's an ellipse, but it's really something else.

 

[rock.freak667]

In the broadest sense, it's an ellipse, but it's really something else.

an ellipse with an eccentricity of zero!

 

[TsAmE]

Yeah I know it looks like a part of an ellipse, but I don't get why the area interpreted is 6pi? Unless it comes from the circle area formula some how (pi *r^2) and is a circle?

 

[Mark44]

Why are you and thepatient insisting the figure is an ellipse?

 

[thepatient]

Why are you and thepatient insisting the figure is an ellipse?

Ohh no, it just looks like half of a slightly tilted ellipse from -2<x<2, but it isn't an ellipse, it looks like a upside down parabola that was stricken by a line coming from the left and merged onto it, though it's domain is only from -2<x<2.

What is kind of strange is that by using the formula for the area of an ellipse and dividing by two, using a=2 and b = 6, abpi/2 = 6pi. But I don't think this is the proper way to calculate this since the end of the parabola isn't at y=6, but at 6.96 to the nearest hundredth and when x = -3/4+(1/4)*41^(1/2).

Either way, integrating it directly gets you 6pi. XD I tried it myself.

 

[Cyosis]

Ohh no, it just looks like half of a slightly tilted ellipse from -2<x<2, but it isn't an ellipse, it looks like a upside down parabola that was stricken by a line coming from the left and merged onto it, though it's domain is only from -2<x<2.

What is kind of strange is that by using the formula for the area of an ellipse and dividing by two, using a=2 and b = 6, abpi/2 = 6pi. But I don't think this is the proper way to calculate this since the end of the parabola isn't at y=6, but at 6.96 to the nearest hundredth and when x = -3/4+(1/4)*41^(1/2).

Either way, integrating it directly gets you 6pi. XD I tried it myself.

Wrong again. Where did you get the values for a and b from? Secondly I suggest you read post #6 and #7 very carefully and try to understand what they are actually trying to tell you.

 

[Mark44]

No, it's definitely not a parabola. Technically, the figure is a kind of ellipse, but that is irrelevant to this problem. I am assuming that we're talking about splitting up the original definite integral into two separate integrals. The first can be done by an ordinary substitution, and the other by using the suggestion given in the problem.

My guess is that TsAmE has not yet been exposed to the integration technique of trig substitution. However, by recognizing that this integral --
$$\int_{-2}^2 \sqrt{4 - x^2} dx$$

-- represents the area of a well-known geometric figure, TsAmE should be able to arrive at the value of the definite integral.

 

[thepatient]

Wrong again. Where did you get the values for a and b from? Secondly I suggest you read post #6 and #7 very carefully and try to understand what they are actually trying to tell you.

Sorry, it's just that I went to graphing directly using the function given and got:

That's why I said tilted parabola. But I see that after you split the integral you just get a simple u-sub and the area of a simple shape.


OP ignore everything I say, hopefully I'm not confusing you. XD I completely understand what mark is saying.

 

[TsAmE]

No, it's definitely not a parabola. Technically, the figure is a kind of ellipse, but that is irrelevant to this problem. I am assuming that we're talking about splitting up the original definite integral into two separate integrals. The first can be done by an ordinary substitution, and the other by using the suggestion given in the problem.

My guess is that TsAmE has not yet been exposed to the integration technique of trig substitution. However, by recognizing that this integral --
$$\int_{-2}^2 \sqrt{4 - x^2} dx$$

-- represents the area of a well-known geometric figure, TsAmE should be able to arrive at the value of the definite integral.

Yeah I haven't done trig substitution yet. I know how to substitute, but how does that get you 6 pi? Since your lower and upper limits are -2 and 2 respectively? If they contained radians, then that would make sense.

 

[Mark44]

This integral represents the area of a common geometric figure (not an ellipse or parabola). What's the figure, and what's its area?

 

[TsAmE]

Not sure, but if I say y = (4 - x^2)^1/2, then I get x^2 + y^4 = 4, which is a circle with radius (2)^1/2

 

[Mark44]

Almost. The equation y = (4 - x^2)^(1/2) represents only part of a circle. The equation x^2 + y^2 = 4 represents an entire circle.

 

[TsAmE]

Makes sense. Since it is 1/4 of a circle, the area equal:

1/4 * pi r^2
= 1/4 * pi * 2
= 1.57

 

[rock.freak667]

Makes sense. Since it is 1/4 of a circle, the area equal:

1/4 * pi r^2
= 1/4 * pi * 2
= 1.57

(4-x²)^1/2 between 0 and 2 give 1/4 of the circle, so between -2 and 2, will be twice the area.

 

[TsAmE]

(4-x²)^1/2 between 0 and 2 give 1/4 of the circle, so between -2 and 2, will be twice the area.

True that, but then:

Area = 1/2 * pi r^2
= 1/2 * pi * 2
= pi

but not 6pi?

 

[Cyosis]

In the original problem that integral is multiplied by (x+3). Secondly the radius of the circle in question is not $\sqrt{2}$.

 

[TsAmE]

In the original problem that integral is multiplied by (x+3). Secondly the radius of the circle in question is not $\sqrt{2}$.

Yeah sorry made a mistake the radius is 2, so now:

Area = 1/2 * pi r^2
= 1/2 * pi * 4
= 6.28

but what does (x + 3) do to the area of (4 - x^2)^1/2 ?

 

[Cyosis]

but what does (x + 3) do to the area of (4 - x^2)^1/2

It does nothing to the area of (4 - x^2)^1/2, but you're not asked to compute the integral of (4 - x^2)^1/2, but the integral of (x+3)(4 - x^2)^1/2. I also suggest you stop fixating on the answer in the back of your book. While it is the correct answer you are now trying to get 6 pi out of everything you do. For example the area underneath the curve (4 - x^2)^1/2 between -2 and 2 is not 6pi. Luckily it isn't because if it was you would not be able to get 6pi as a final answer.

 

[Mentallic]

A circle of radius 2, not $\sqrt{2}$.

I'm not quite sure what this shape is either. Just tell us already Mark! It's the addition of a circle and a sideways S. They have a name for this sort of thing?

As to TsAmE, not sure why you're tackling this problem when you don't know trig substitution yet.

 

[Cyosis]

I'm not quite sure what this shape is either. Just tell us already Mark! It's the addition of a circle and a sideways S. They have a name for this sort of thing?

As to TsAmE, not sure why you're tackling this problem when you don't know trig substitution yet.

People have been talking about the shape of $(4-x^2)^{1/2}$, which is part of a circle as TsAmE figured out already. The full integrand's shape does not have a special name, not that I am aware of anyway.

You really don't want to evaluate this integral by substitution, whether you're able to or not. Realising what the integral represents allows you to instantly write down the answer.

 

[thepatient]

Area of a circle is pi*r^2, not 1/2 * pi * r^2. And you have split the integrand into x(4-x^2)^1/2 +3(4-x^2)^1/2. One of those becomes zero since it's a function symmetric to the origin, and one is just the area over a circle.

 

[Cyosis]

Area of a circle is pi*r^2, not 1/2 * pi * r^2. And you have split the integrand into x(4-x^2)^1/2 +3(4-x^2)^1/2. One of those becomes zero since it's a function symmetric to the origin, and one is just the area over a circle.

The area beneath $(4-x^2)^{1/2}$ between -2 and 2 is half the area of a circle. Therefore TsAmE's calculation of the area is correct.

 

[thepatient]

The area beneath $(4-x^2)^{1/2}$ between -2 and 2 is half the area of a circle. Therefore TsAmE's calculation of the area is correct.

Oh yea. Half circle. XD I'm so asleep still.

 

[HallsofIvy]

ThePatient's graph shows only part of the figure and has the axes with very different unit sizes.

 

[SyNtHeSiS]

It does nothing to the area of (4 - x^2)^1/2, but you're not asked to compute the integral of (4 - x^2)^1/2, but the integral of (x+3)(4 - x^2)^1/2. I also suggest you stop fixating on the answer in the back of your book. While it is the correct answer you are now trying to get 6 pi out of everything you do. For example the area underneath the curve (4 - x^2)^1/2 between -2 and 2 is not 6pi. Luckily it isn't because if it was you would not be able to get 6pi as a final answer.

The question asked to interpret it in terms of area, that's why I didnt use any integration technique e.g. substitution, but instead tried to calculate in terms of half a circle (1/2 * pi * r^2). So is the answer at the back of my book wrong?