I am confused on how to solve this. NOTE: the lower limit for the second function is (-6) NOT 6. Wouldn't let me type it in correctly.[tex]\int\limits^6_3 {|x-4|} \, dx + \int\limits^0_6 {\sqrt{36-x^2} } \, dx[/tex]
Answers
I think how you approach this depends on your knowledge of calculus.
If you don't know how to compute definite integrals yet, but you do know that they represent signed areas under curves, then you can plot both curves |x - 4| and √(36 - x ²), then recognize that the areas represented by these integrals are areas of geometric shapes. (See attached images)
First integral: if you plot |x - 4| on the interval [3, 6], you'll see that the integral corresponds to the area of two triangles. One of them has base = height = 1, and the other has base = height = 2. Then
[tex]\displaystyle\int_3^6|x-4|\,\mathrm dx=\frac12\times1\times1+\frac12\times2\times2=\frac52[/tex]
Second integral: if [tex]y=\sqrt{36-x^2}[/tex], then [tex]x^2+y^2=6^2[/tex], meaning this curve is the upper half of a circle with radius 6. On the interval [-6, 0], the area amounts to 1/4 of the total area of such a circle, so that
[tex]\displaystyle\int_{-6}^0\sqrt{36-x^2}\,\mathrm dx=\frac{\pi\times6^2}4=9\pi[/tex]
* * *
If you already know a few things about calculus and integration, you can compute these areas directly.
First integral:
[tex]\displaystyle\int_3^6|x-4|\,\mathrm dx=\int_3^4(4-x)\,\mathrm dx+\int_4^6(x-4)\,\mathrm dx[/tex]
[tex]\displaystyle\int_3^6|x-4|\,\mathrm dx=\left(4x-\frac{x^2}2\right)\bigg|_{x=3}^{x=4}+\left(\frac{x^2}2-4x\right)\bigg|_{x=4}^{x=6}[/tex]
[tex]\displaystyle\int_3^6|x-4|\,\mathrm dx=\left(\left(4\times4-\frac{4^2}2\right)-\left(4\times3-\frac{3^2}2\right)\right)+\left(\left(\frac{6^2}2-4\times6\right)-\left(\frac{4^2}2-4\times4\right)\right)[/tex]
[tex]\displaystyle\int_3^6|x-4|\,\mathrm dx=\frac52[/tex]
Second integral:
Substitute x = 6 sin(t ) and dx = 6 cos(t ) dt, then
[tex]\displaystyle\int_{-6}^0\sqrt{36-x^2}\,\mathrm dx=\int_{-\frac\pi2}^0\sqrt{6^2-(6\sin(t))^2} (6\cos(t))\,\mathrm dt[/tex]
[tex]\displaystyle\int_{-6}^0\sqrt{36-x^2}\,\mathrm dx=36\int_{-\frac\pi2}^0\cos (t) \sqrt{1-\sin^2(t)} \,\mathrm dt[/tex]
[tex]\displaystyle\int_{-6}^0\sqrt{36-x^2}\,\mathrm dx=36\int_{-\frac\pi2}^0\cos (t) \sqrt{\cos^2(t)} \,\mathrm dt[/tex]
[tex]\displaystyle\int_{-6}^0\sqrt{36-x^2}\,\mathrm dx=36\int_{-\frac\pi2}^0\cos (t) |\cos(t)| \,\mathrm dt[/tex]
For t ∈ [-π/2, 0], cos(t ) > 0, so |cos(t )| = cos(t ) :
[tex]\displaystyle\int_{-6}^0\sqrt{36-x^2}\,\mathrm dx=36\int_{-\frac\pi2}^0\cos^2(t)\,\mathrm dt[/tex]
Recall the half-angle identity,
cos²(t ) = (1 + cos(2t )) / 2
[tex]\displaystyle\int_{-6}^0\sqrt{36-x^2}\,\mathrm dx=18\int_{-\frac\pi2}^0(1+\cos(2t))\,\mathrm dt[/tex]
[tex]\displaystyle\int_{-6}^0\sqrt{36-x^2}\,\mathrm dx=18\left(t+\frac{\sin(2t)}2\right)\bigg|_{t=-\frac\pi2}^{t=0}[/tex]
[tex]\displaystyle\int_{-6}^0\sqrt{36-x^2}\,\mathrm dx=18\left(\left(0+\frac{\sin(2\times0)}2\right)-\left(-\frac\pi2+\frac{\sin\left(2\times\left(-\frac\pi2\right)\right)}2\right)\right)[/tex]
[tex]\displaystyle\int_{-6}^0\sqrt{36-x^2}\,\mathrm dx=9\pi[/tex]
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Answer:
43.5%
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Answers
Answer:
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Step-by-step explanation:
arithmetic sequences are in form:
a(n)=a(o)+d(n-1)
sum are in form:
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Answers
Answer:
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Answer:
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Answer:
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Step-by-step explanation:
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Answers
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Step-by-step explanation:
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Answers
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Answer:
Why does this table start at zero?The vertical axis on trending charts should almost always start at zero. The reason it's almost always is because you may have trending data that contains negative values or fractions. In those situations, it's generally best to keep Excel's default scaling.
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Step-by-step explanation:
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No
Step-by-step explanation:
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Answer:
He can be 99% confident that the mean number of calls per day is between 266.5 and 292.3.
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Upper bound: x calls
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Answer:
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Step-by-step explanation:
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May someone please explain how I would reflect it?
Answers
R’ (2, -3)
X’ (-2, 0)
Z’ (-2, -2)
EXPLANATION:
Reflecting a figure across the y-axis basically means flipping all of the x-values of your coordinates while retaining its y-value.
In other words, flip the signs of x and keep y the exact same for each coordinate.
B (-3, 2)
=
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Answer:
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Step-by-step explanation:
I can not interfere with your destiny you will need to decide your destiny
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the sumof the present ages of a father and his son is 60 years. 6years ago, the father's age was 5 times the age of his son.after 6 years, how old will the son be?
Answers
Answer:
20 years
Step-by-step explanation:
Let the present ages of son and father be x and (60 -x) years respectively.
Then, (60 - x) - 6 = 5(x - 6)
54 - x = 5x - 30
6x = 84
x = 14.
Son's age after 6 years = (x+ 6) = 20 years..
I hope this helps a little bit.