2022-05-01

Find the area of the region enclosed by one loop of the curve v=3sin2θ

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Skilled2022-05-13Added 403 answers

${\int}_{0}^{\frac{\pi}{2}}\frac{1}{2}\left(3{\mathrm{sin}}^{2}\left(2\theta \right)\right)d\theta$

Write the problem as a mathematical expression.

${\int}_{0}^{\frac{\pi}{2}}\frac{1}{2}\left(3{\mathrm{sin}}^{2}\left(2\theta \right)\right)d\theta$

Simplify.

${\int}_{0}^{\frac{\pi}{2}}\frac{3{\mathrm{sin}}^{2}\left(2\theta \right)}{2}d\theta$

Since $\frac{3}{2}$ is constant with respect to $\theta$, move $\frac{3}{2}$ out of the integral.

$\frac{3}{2}{\int}_{0}^{\frac{\pi}{2}}{\mathrm{sin}}^{2}\left(2\theta \right)d\theta$

Let ${u}_{1}=2\theta$. Then $d{u}_{1}=2d\theta$, so $\frac{1}{2}d{u}_{1}=d\theta$. Rewrite using $u}_{1$ and $d$$u}_{1$.

$\frac{3}{2}{\int}_{0}^{\pi}{\mathrm{sin}}^{2}\left({u}_{1}\right)\frac{1}{2}d{u}_{1}$

Combine ${\mathrm{sin}}^{2}\left({u}_{1}\right)$ and $\frac{1}{2}$.

$\frac{3}{2}{\int}_{0}^{\pi}\frac{{\mathrm{sin}}^{2}\left({u}_{1}\right)}{2}d{u}_{1}$

Since $\frac{1}{2}$ is constant with respect to $u}_{1$, move $\frac{1}{2}$ out of the integral.

$\frac{3}{2}\left(\frac{1}{2}{\int}_{0}^{\pi}{\mathrm{sin}}^{2}\left({u}_{1}\right)d{u}_{1}\right)$

Simplify.

Multiply $\frac{1}{2}$ by $\frac{3}{2}$.

$\frac{3}{2\cdot 2}{\int}_{0}^{\pi}{\mathrm{sin}}^{2}\left({u}_{1}\right)d{u}_{1}$

Multiply $2$ by $2$.

$\frac{3}{4}{\int}_{0}^{\pi}{\mathrm{sin}}^{2}\left({u}_{1}\right)d{u}_{1}$

Use the half-angle formula to rewrite ${\mathrm{sin}}^{2}\left({u}_{1}\right)$ as $\frac{1-\mathrm{cos}\left(2{u}_{1}\right)}{2}$.

$\frac{3}{4}{\int}_{0}^{\pi}\frac{1-\mathrm{cos}\left(2{u}_{1}\right)}{2}d{u}_{1}$

Since $\frac{1}{2}$ is constant with respect to $u}_{1$, move $\frac{1}{2}$ out of the integral.

$\frac{3}{4}(\frac{1}{2}{\int}_{0}^{\pi}1-\mathrm{cos}\left(2{u}_{1}\right)d{u}_{1})$

Simplify.

Multiply $\frac{1}{2}$ by $\frac{3}{4}$.

$\frac{3}{2\cdot 4}{\int}_{0}^{\pi}1-\mathrm{cos}\left(2{u}_{1}\right)d{u}_{1}$

Multiply $2$ by $4$.

$\frac{3}{8}{\int}_{0}^{\pi}1-\mathrm{cos}\left(2{u}_{1}\right)d{u}_{1}$

Split the single integral into multiple integrals.

$\frac{3}{8}({\int}_{0}^{\pi}d{u}_{1}+{\int}_{0}^{\pi}-\mathrm{cos}\left(2{u}_{1}\right)d{u}_{1})$

Apply the constant rule.

$\frac{3}{8}({u}_{1}{]}_{0}^{\pi}+{\int}_{0}^{\pi}-\mathrm{cos}\left(2{u}_{1}\right)d{u}_{1})$

Since $-1$ is constant with respect to $u}_{1$, move $-1$ out of the integral.

$\frac{3}{8}({u}_{1}{]}_{0}^{\pi}-{\int}_{0}^{\pi}\mathrm{cos}\left(2{u}_{1}\right)d{u}_{1})$

Let $u}_{2}=2{u}_{1$. Then $d{u}_{2}=2d{u}_{1}$, so $\frac{1}{2}d{u}_{2}=d{u}_{1}$. Rewrite using $u}_{2$ and $d$$u}_{2$.

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$\frac{3}{8}({u}_{1}{]}_{0}^{\pi}-{\int}_{0}^{2\pi}\mathrm{cos}\left({u}_{2}\right)\frac{1}{2}d{u}_{2})$

Combine $\mathrm{cos}\left({u}_{2}\right)$ and $\frac{1}{2}$.

$\frac{3}{8}({u}_{1}{]}_{0}^{\pi}-{\int}_{0}^{2\pi}\frac{\mathrm{cos}\left({u}_{2}\right)}{2}d{u}_{2})$

Since $\frac{1}{2}$ is constant with respect to $u}_{2$, move $\frac{1}{2}$ out of the integral.

$\frac{3}{8}({u}_{1}{]}_{0}^{\pi}-\left(\frac{1}{2}{\int}_{0}^{2\pi}\mathrm{cos}\left({u}_{2}\right)d{u}_{2}\right))$

The integral of $\mathrm{cos}\left({u}_{2}\right)$ with respect to $u}_{2$ is $\mathrm{sin}\left({u}_{2}\right)$.

$\frac{3}{8}({u}_{1}{]}_{0}^{\pi}-\frac{1}{2}\mathrm{sin}\left({u}_{2}\right){]}_{0}^{2\pi})$

Substitute and simplify.

$\frac{3}{8}(\pi -\frac{1}{2}(\mathrm{sin}\left(2\pi \right)-\mathrm{sin}\left(0\right)))$

Simplify.

$\frac{3}{8}(\pi -\frac{\mathrm{sin}\left(2\pi \right)}{2})$

Simplify.

$\frac{3\pi}{8}$

The result can be shown in multiple forms.

$\frac{3\pi}{8}$

(x + 3y)dx – xdy = 0

2xdy/dx=1+y^2

$2xdy/dx=1+y^2$2xdy/dx=1+y^2

Find values of x, if any, at which f is not continuous.

f(x) = 5x^4 − 3x + 7

determine the following functions (v/g)x

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a. How high is the arch at its center approximated in two decimal places?

b. What is the horizontal length, approximated to two decimal places, of the arch from axis when it’s 15 feet high?

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4x x 3x -4 +19

find the derivative of g(x)= x^2-4/x+0.5 by quotient rule

How do you find a Least Common Factor in cases like

Example:

2/5 x 5/7

The table that is needed has been provided in the images.

Given the differential $=y{}^{\u2033}-8{y}^{\prime}+16y=-3{e}^{4x}$.

a) Find the complementary solution $={y}_{c}$ to the differential equation.

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$={y}_{p}=$__________________________Test x+y=9

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a. If each futures contract costs $25 to trade with a discount broker, how much is the transaction cost per dollar of stock controlled by the futures contract?

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Photographers often use multiple sources of light light

Control the shadows or to enlighten their subject in an artistic way. The light

from a light source is inversely proportional to the Carrá

The distance from the source and proportional has its intensity. Two Sone

of light, Li and L2, are placed 10 m away, and have respectively

An intensity of 4 units and 1 unit. Leclairage in any point P

corresponds to the sum of the lighting from the two sources.

An intense light source

Hint: when a quantity is inversely proportional to the square of a, where K is a constant.

a) Determine a function which represents the lighting in point P according to its

Distance from point L

b) How much distance is the lighting the most powerful in P?

) At what distance from the point is the lighting the weakest in P?

d) examine whether it would be more effective to increase the intensity of L2 or to place

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Press your answer using mathematical evidence. Take into account the

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or the position of L2 varies.