Mathematics, 10.03.2021 01:00 PKTHUNDER5011

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. $\int\limits^6_3 {|x-4|} \, dx + \int\limits^0_6 {\sqrt{36-x^2} } \, dx$

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Answer from: pillgiing

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

$\displaystyle\int_3^6|x-4|\,\mathrm dx=\frac12\times1\times1+\frac12\times2\times2=\frac52$

Second integral: if $y=\sqrt{36-x^2}$ , then x^2+y^2=6^2 , 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

$\displaystyle\int_{-6}^0\sqrt{36-x^2}\,\mathrm dx=\frac{\pi\times6^2}4=9\pi$

* * *

If you already know a few things about calculus and integration, you can compute these areas directly.

First integral:

$\displaystyle\int_3^6|x-4|\,\mathrm dx=\int_3^4(4-x)\,\mathrm dx+\int_4^6(x-4)\,\mathrm dx$

$\displaystyle\int_3^6|x-4|\,\mathrm dx=\left(4x-\frac{x^2}2\right)\bigg|_{x=3}^{x=4}+\left(\frac{x^2}2-4x\right)\bigg|_{x=4}^{x=6}$

$\displaystyle\int_3^6|x-4|\,\mathrm dx=\left(\left(4\times4-\frac{4^2}2\right)-\left(4\times3-\frac{3^2}2\right)\right)+\left(\left(\frac{6^2}2-4\times6\right)-\left(\frac{4^2}2-4\times4\right)\right)$

$\displaystyle\int_3^6|x-4|\,\mathrm dx=\frac52$

Second integral:

Substitute x = 6 sin(t ) and dx = 6 cos(t ) dt, then

$\displaystyle\int_{-6}^0\sqrt{36-x^2}\,\mathrm dx=\int_{-\frac\pi2}^0\sqrt{6^2-(6\sin(t))^2} (6\cos(t))\,\mathrm dt$

$\displaystyle\int_{-6}^0\sqrt{36-x^2}\,\mathrm dx=36\int_{-\frac\pi2}^0\cos (t) \sqrt{1-\sin^2(t)} \,\mathrm dt$

$\displaystyle\int_{-6}^0\sqrt{36-x^2}\,\mathrm dx=36\int_{-\frac\pi2}^0\cos (t) \sqrt{\cos^2(t)} \,\mathrm dt$

$\displaystyle\int_{-6}^0\sqrt{36-x^2}\,\mathrm dx=36\int_{-\frac\pi2}^0\cos (t) |\cos(t)| \,\mathrm dt$

For t ∈ [-π/2, 0], cos(t ) > 0, so |cos(t )| = cos(t ) :

$\displaystyle\int_{-6}^0\sqrt{36-x^2}\,\mathrm dx=36\int_{-\frac\pi2}^0\cos^2(t)\,\mathrm dt$

Recall the half-angle identity,

cos²(t ) = (1 + cos(2t )) / 2

$\displaystyle\int_{-6}^0\sqrt{36-x^2}\,\mathrm dx=18\int_{-\frac\pi2}^0(1+\cos(2t))\,\mathrm dt$

$\displaystyle\int_{-6}^0\sqrt{36-x^2}\,\mathrm dx=18\left(t+\frac{\sin(2t)}2\right)\bigg|_{t=-\frac\pi2}^{t=0}$

$\displaystyle\int_{-6}^0\sqrt{36-x^2}\,\mathrm dx=18\left(\left(0+\frac{\sin(2\times0)}2\right)-\left(-\frac\pi2+\frac{\sin\left(2\times\left(-\frac\pi2\right)\right)}2\right)\right)$

$\displaystyle\int_{-6}^0\sqrt{36-x^2}\,\mathrm dx=9\pi$

I am confused on how to solve this. NOTE: the lower limit for the second function is (-6) NOT 6. Wou

Answer from: Quest

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.

Answer from: Quest

do your own work and don"t cheat.

step-by-step explanation:

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