Using area to evaluate integrals

[Mhorton91]

Homework Statement

Use areas to evaluate the integral f(x)=5x+√(25-x2) on the following intervals

a) [-5,0]
b) [-5,5]

Homework Equations

∫f(x) + g(x) = ∫f(x) + ∫g(x)

also Area of a circle = pi(r)²

The Attempt at a Solution

[/B]
My first several attempts have centered around evaluating the integral on the first interval, from -5 to 0.

I have attempted to break it into---

∫5x + ∫√(25-x².

After graphing y=5x, I get a point at (-5, -25) which I use to construct a triangle of base (-)5, and height (-)25.

Using the 1/2(base)(height) formula I get 125/2 square units.

The graph of the latter y=√(25-x² appears to be the upper hemisphere of a circle of radius 5, centered around the origin.

Limiting it to the interval [-5,0] Leaves me with 1/4 of a circle radius 5, or

1/4(pi)(radius²)
To which I end up getting 25pi/4

so then I attempt to combine them with the rule of sums listed above.. My class uses an online homework system, and the answer I am entering is

(25pi/4)+125/2

That answer doesn't appear to be correct.

I had to take a year off after Calc 1, and this is my first assignment in Calc 2.. I'm trying to get back on the learning curve.Also, here is a screen shot of the actual problem in case my write up was confusing.

I appreciate any push in the proper direction!

Marshall

 

[RUber]

Could the computer be asking for a decimal answer, like 6.25 pi + 62.5?

 

[RUber]

Oh, I see now. Your triangle is negative area--You have to subtract it.
The integral
$\int_{-5}^0 5x dx = \frac52 x^2 |_{-5}^0 = \frac52 (0)^2 - \frac52 (-5)^2 = -125/2.$
Area below the x-axis will result in a negative integral.
Keep that in mind when you add the second part of the interval on from [0,5] where both functions are positive over the interval.

 

[Ray Vickson]

Homework Statement

Use areas to evaluate the integral f(x)=5x+√(25-x2) on the following intervals

a) [-5,0]
b) [-5,5]

Homework Equations

∫f(x) + g(x) = ∫f(x) + ∫g(x)

also Area of a circle = pi(r)²

The Attempt at a Solution

[/B]
My first several attempts have centered around evaluating the integral on the first interval, from -5 to 0.

I have attempted to break it into---

∫5x + ∫√(25-x².

After graphing y=5x, I get a point at (-5, -25) which I use to construct a triangle of base (-)5, and height (-)25.

Using the 1/2(base)(height) formula I get 125/2 square units.

The graph of the latter y=√(25-x² appears to be the upper hemisphere of a circle of radius 5, centered around the origin.

Limiting it to the interval [-5,0] Leaves me with 1/4 of a circle radius 5, or

1/4(pi)(radius²)
To which I end up getting 25pi/4

so then I attempt to combine them with the rule of sums listed above.. My class uses an online homework system, and the answer I am entering is

(25pi/4)+125/2

That answer doesn't appear to be correct.

I had to take a year off after Calc 1, and this is my first assignment in Calc 2.. I'm trying to get back on the learning curve.I appreciate any push in the proper direction!

Marshall

Are familiar with the fundamental theorem of (integral) calculus? It says that for a function $f(x)$ the area under the curve $y = f(x)$ from $x = a$ to $x = b$ is
$$\text{Area} = F(b) - F(a),$$
where $F(x)$ is an antiderivative function of $f(x)$; that is, it is a function whose derivative $F'(x)$ equals $f(x)$. So, you need only find the antiderivatives of the two functions $5x$ and $\sqrt{25 - x^2}$. You need not bother with trying to construct triangles and semi-circles; just do integrations. You should have done both of these types of integrations in Calculus 1, although you may need to review that material.

Be careful, though: "area under the curve" can be negative, as it would be for the first term $5x$ on the interval $-5 \leq x \leq 0$, and the whole curve $y = 5x + \sqrt{25 - x^2}$ for part of the interval.

 

[SammyS]

Homework Statement

Use areas to evaluate the integral f(x)=5x+√(25-x²) on the following intervals

a) [-5,0]
b) [-5,5]

Homework Equations

∫f(x) + g(x) = ∫f(x) + ∫g(x)

also Area of a circle = pi(r)²

The Attempt at a Solution

[/B]
My first several attempts have centered around evaluating the integral on the first interval, from -5 to 0.

[ IMG]https://lh5.googleusercontent.com/K...DRLAHbyN722kWKzoj8GHlBugKZLynfMXD7CAw[/PLAIN]

I appreciate any push in the proper direction!

Marshall

It's probably easier to work with the interval [-5, 5] first.

 

[Mhorton91]

Alright, quote function isn't wanting to work, so go with me here.

RUber. The triangle has a negative area.. That definitely makes sense now that I look at it.

So I'd be looking at something like

(25pi/4) - (125/2)?

I'm at work right now, and can't sit down and rework the problem with that prompt, so that is just the intuitive response. Thank you.

Ray Vickson- I was familiar with the fundamental theorem of integral calculus, and I do remember learning how to find anti derivatives to solve integrals.. I also remember being fairly comfortable with it.. I guess I need to go back to my Calc 1 notebook. Thank you.

SammyS- I had that thought, however the software prompts for the interval [-5,0] and requires an answer prior to prompting for the second interval. Thanks everyone for the help! I hate being behind the curve!
Marshall.Side question, is it my phone, or just how the app is made that it won't show the formulas you have entered?

 

[RUber]

RUber. The triangle has a negative area.. That definitely makes sense now that I look at it.

So I'd be looking at something like

(25pi/4) - (125/2)?

I'm at work right now, and can't sit down and rework the problem with that prompt, so that is just the intuitive response. Thank you.

That should be right.

 

[Mhorton91]

That should be right.

Thanks!

I definitely need to track down my Calc 1 notebook and start working my way back through.

I guess this is a lesson to not take a year off again.

 

[RUber]

Thanks!

I definitely need to track down my Calc 1 notebook and start working my way back through.

I guess this is a lesson to not take a year off again.

We all need some time off now and then. I took 7 years off and jumped back into calc 3. It wasn't pretty, but only a few weeks in it got a lot better.

 

[Mhorton91]

We all need some time off now and then. I took 7 years off and jumped back into calc 3. It wasn't pretty, but only a few weeks in it got a lot better.

I'm hoping for something similar.

This is my first semester at this university. Prior I was at a community college, however their math department was solid, and our Calc 1 class went far enough in that according to our course schedule I've already been exposed to about 2.5 weeks of material... I'm hoping by the time I get to new material I will have transitioned back into the academic mindset.

Thanks again for the help everyone,
Marshall

 

[almassing]

Compute the definite integral:
integral_(-5)^0 (sqrt(25-x^2)+5 x) dx
Integrate the sum term by term and factor out constants:
= integral_(-5)^0 sqrt(25-x^2) dx+5 integral_(-5)^0 x dx
For the integrand sqrt(25-x^2), substitute x = 5 sin(u) and dx = 5 cos(u) du.
Then sqrt(25-x^2) = sqrt(25-25 sin^2(u)) = 5 sqrt(cos^2(u)).
This substitution is invertible over -pi/2<u<0 with inverse u = sin^(-1)(x/5).
This gives a new lower bound u = sin^(-1)(-5/5) = -pi/2 and upper bound u = sin^(-1)(0/5) = 0:
= 5 integral_(-pi/2)^0 5 cos(u) sqrt(cos^2(u)) du+5 integral_(-5)^0 x dx
Factor out constants:
= 25 integral_(-pi/2)^0 cos(u) sqrt(cos^2(u)) du+5 integral_(-5)^0 x dx
Simplify cos(u) sqrt(cos^2(u)) assuming -pi/2<u<0:
= 25 integral_(-pi/2)^0 cos^2(u) du+5 integral_(-5)^0 x dx
Write cos^2(u) as 1/2 cos(2 u)+1/2:
= 25 integral_(-pi/2)^0 (1/2 cos(2 u)+1/2) du+5 integral_(-5)^0 x dx
Integrate the sum term by term and factor out constants:
= 25/2 integral_(-pi/2)^0 cos(2 u) du+25/2 integral_(-pi/2)^0 1 du+5 integral_(-5)^0 x dx
For the integrand cos(2 u), substitute s = 2 u and ds = 2 du.
This gives a new lower bound s = (2 (-pi))/2 = -pi and upper bound s = 2 0 = 0:
= 25/4 integral_(-pi)^0 cos(s) ds+25/2 integral_(-pi/2)^0 1 du+5 integral_(-5)^0 x dx
Apply the fundamental theorem of calculus.
The antiderivative of cos(s) is sin(s):
= (25 sin(s))/4|_(-pi)^0+25/2 integral_(-pi/2)^0 1 du+5 integral_(-5)^0 x dx
Evaluate the antiderivative at the limits and subtract.
(25 sin(s))/4|_(-pi)^0 = (25 sin(0))/4-(25 sin(-pi))/4 = 0:
= 25/2 integral_(-pi/2)^0 1 du+5 integral_(-5)^0 x dx
Apply the fundamental theorem of calculus.
The antiderivative of 1 is u:
= (25 u)/2|_(-pi/2)^0+5 integral_(-5)^0 x dx
Evaluate the antiderivative at the limits and subtract.
(25 u)/2|_(-pi/2)^0 = (25 0)/2-((25 (-pi))/(2 2)) = (25 pi)/4:
= (25 pi)/4+5 integral_(-5)^0 x dx
Apply the fundamental theorem of calculus.
The antiderivative of x is x^2/2:
= (25 pi)/4+(5 x^2)/2|_(-5)^0
Evaluate the antiderivative at the limits and subtract.
(5 x^2)/2|_(-5)^0 = (5 0^2)/2-(5 (-5)^2)/2 = -125/2:
= (25 pi)/4-125/2
Which is equal to:
Answer: |
| = 25/4 (pi-10)