Sketch the region enclosed by the curves and find its area

[warwick]

9. y=cos(2x), y=0, x=pi/4, x=pi/2

I got 1/2 for number 9, and I just discovered my original substitution error. Fortunately, though, I was integrating the constant, 0, which incidentally did not affect my computation. I still want to make sure I did it correctly. I integrated "0x" with respect to x from (pi/4)to (pi/2) and sin u with respect to u from (pi/2) to (pi). Mainly, I need some quick review lessons on integrating trigonometric functions, exponentials, logarithms, and especially doing it with substitution, etc. It's starting to slowly come back to me, but unfortunately it's during my first Cal II homework set. Haha.

13. y=e^x, y=e^2x, x=0, x=ln 2

Ok. I broke this up into two definite integrals because one needed substitution, the other did not. My answer is 1/2.

15. y=(2)/(1+x^2), y=absolute value of x

Maybe use the Trapezoidal Rule here?

17. y=2+ absolute value of (x-1), y= (-1/5x) +7

 

[galactus]

Re: Sketch the region enclosed by the curves and find its ar

17.

\(\displaystyle \L\\y=2+ |x-1|, \;\ y= (\frac{-1}{5}x) +7\)

\(\displaystyle \L\\|x-1|=(x-1) \;\ and \;\ -(x-1)\)

Whenever you see an absolute value to integrate, try breaking it up into two integrals.

\(\displaystyle \L\\\int_{-5}^{1}[\frac{-1}{5}x+7-(2-(x-1))]dx+\int_{1}^{5}[\frac{-1}{5}x+7-(2+(x-1))]dx\)

 

[soroban]

Re: Sketch the region enclosed by the curves and find its ar

Hello, warwick!

\(\displaystyle 9)\; y\,=\,\cos(2x),\;y\,=\,0,\;x\,=\,\frac{\pi}{4},\; x\,=\,\frac{\pi}{2}\)

If the lower boundary is the x-axis, we don't have to include it in the integral.

\(\displaystyle \L A \;=\;\int^{\;\;\;\frac{\pi}{2}}_{\frac{\pi}{4}}\cos(2x)\,dx \;=\;\frac{1}{2}\sin(2x)\,\bigg]^{\frac{\pi}{2}}_{\frac{\pi}{4}}\;\;\) . . . etc.

\(\displaystyle 13)\;y\,=\,e^x,\;\;y\,=\,e^{2x},\;\;x\,=\,0,\;\;x\,=\,\ln2\)

\(\displaystyle \L A \;=\;\int^{\;\;\;\;\;\ln2}_0\left(e^{2x}\,-\,e^x\right)\,dx \;=\;\frac{1}{2}e^{2x}\,-\,e^x\,\bigg]^{\ln2}_0\;\;\) . . . etc.

Your answer is correct!

\(\displaystyle 15)\; y\:=\:\frac{2}{1\,+\,x^2},\;\;y\,=\,|x|\)

Did you make a sketch?

                    |
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          \     *:::|:::*     /
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             *  \:::|:::/ :*
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      - - - - - - - * - - + - - - - -
                    |     1

The area is: \(\displaystyle \L\:A \;=\;2\,\times\int^{\;\;\;1}_0\left(\frac{2}{1+x^2}\,-\,x\right)\,dx \;=\;2\left[2\arctan x \,- \,\frac{1}{2}x^2\,\right]^1_0\)

\(\displaystyle 17)\;[1]\;y\:=\:2\,+\,|x\,-\,1|,\;\;[2]\;y\:=\:-\frac{1}{5}x\,+\,7\)

The graph of \(\displaystyle [1]\) is the graph of \(\displaystyle y\:=\:|x|\)
. . moved 1 unit to the right and 2 units up.

The graph of \(\displaystyle [2]\) is a straight line
. . with y-intercept \(\displaystyle 7\) and slope -\(\displaystyle \frac{1}{5}\)

      *       |
      ::::*   |
      : \ ::::*
      :   \:::|:::*           /
      :     \:|:::::::*     /
      :       |:::::::::::*
      :       | \:::::::/ :   *
      :       |   \:::/   :
      :       |     *     :
      :       |     :     : 
      :       |     :     :
    - + - - - + - - + - - + - - -
     -5       |     1     5

If \(\displaystyle x\,<\,1,\:[1]\) is: \(\displaystyle \,y\:=\:2\,-\,(x\,-\,1)\;\;\Rightarrow\;\;y\:=\:3\,-\,x\)
. . It intersects [2] when: \(\displaystyle \:3\,-\,x\:=\:-\frac{1}{5}x\,+\,7\;\;\Rightarrow\;\;x\,=\,-5\)

If \(\displaystyle x\,>\,1,\:[2]\) is: \(\displaystyle \,y\:=\:2\,+\,(x\,-\,1)\;\;\Rightarrow\;\;y\:=\:x\,+\,1\)
. . It intersects [2] when: \(\displaystyle \:x\,+\,1\:=\:-\frac{1}{5}x\,+\,7\;\;\Rightarrow\;\;x\,=\,5\)

\(\displaystyle \L A \;=\;\int^{\;\;\;1}_{-5}\left[\left(-\frac{1}{5}x\,+\,7\right)\,-\,(3\,-\,x)\right]\,dx\:+\:\int^{\;\;\;5}_1\left[\left(-\frac{1}{5}x\,+\,7\right)\,-\,(x\,+\,1)\right]\,dx\)

 

[warwick]

You're going to have to work out how you got the Area equation in 15. I don't remember that identity at all.

I'll get working on 17 here in a bit.

 

[galactus]

I don't remember that identity at all.

That's not an identity, per se, it's the integral of \(\displaystyle \frac{1}{1+x^{2}}\)

\(\displaystyle \L\\\int\frac{1}{1+x^{2}}dx=tan^{-1}(x)\)

 

[stapel]

You're going to have to work out how you got the Area equation in 15.

Um... isn't the "working out" bit sort of what you're supposed to do...? :shock:

You were given the "how", given the sketch of the "how", given the complete area equation based on the sketch, and given the integrated solution. All that's left is looking at the provided sketch, comparing with the provided equation, and then evaluating the provided solution. :wink:

Where are you stuck? Please be complete. Thank you!

eliz.

 

[warwick]

You're going to have to work out how you got the Area equation in 15.

Um... isn't the "working out" bit sort of what you're supposed to do...? :shock:

You were given the "how", given the sketch of the "how", given the complete area equation based on the sketch, and given the integrated solution. All that's left is looking at the provided sketch, comparing with the provided equation, and then evaluating the provided solution. :wink:

Where are you stuck? Please be complete. Thank you!

eliz.

Ok. I see how the Area formula was derived because of symmetry. I guess I'm not sure how to integrate 1/(1+x^2)

For answer 17 I got 24.

 

[stapel]

I guess I'm not sure how to integrate 1/(1+x^2)

Just use the trig-integration identity the other tutor gave you earlier. (And, if you haven't yet memorized that table of trig- and inverse-trig integrals, do it now!!)

Eliz.

 

[galactus]

These are mostly memorized in lieu of derivation. Look under derivatives and integrals of inverse trig functions in your text (assuming you have one).

Here's one way we can show the derivative of arc tan.

Let \(\displaystyle \L\\y=tan^{-1}(x), \,\ x=tan(y)\)

Differentiate the second one implicitly wrt x gives us:

\(\displaystyle \L\\\frac{d}{dx}[x]=\frac{d}{dx}[tan(y)]\)

\(\displaystyle \L\\1=sec^{2}(y)\frac{dy}{dx}\)

\(\displaystyle \L\\\frac{dy}{dx}=\frac{1}{sec^{2}(y)}=\frac{1}{sec^{2}(tan^{-1}(x))}\)..........[1]

Then we get from Pythagoras (see diagram below)


\(\displaystyle \L\\sec^{2}(tan^{-1}(x))=1+x^{2}\)

And [1] simplifies to : \(\displaystyle \L\\\frac{1}{1+x^{2}}\)

Of course, then \(\displaystyle \L\\\int\frac{1}{1+x^{2}}dx=tan^{-1}(x)\)

 

[warwick]

Hmm. I don't remember seeing those in my Cal I book last year. I had to buy a different book for my Cal II class. Well, that would've made my question unnecessary.