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matrenka [14]
3 years ago
12

(a) Find the Riemann sum for f(x) = 4 sin(x), 0 ≤ x ≤ 3π/2, with six terms, taking the sample points to be right endpoints. (Rou

nd your answers to six decimal places.)

Mathematics
1 answer:
miskamm [114]3 years ago
5 0

Split up the integration interval into 6 subintervals:

\left[0,\dfrac\pi4\right],\left[\dfrac\pi4,\dfrac\pi2\right],\ldots,\left[\dfrac{5\pi}4,\dfrac{3\pi}2\right]

where the right endpoints are given by

r_i=i\dfrac{\frac{3\pi}2-0}6=\dfrac{i\pi}4

for 1\le i\le6. Then we approximate the integral

\displaystyle\int_0^{3\pi/2}4\sin x\,\mathrm dx

by the Riemann sum,

\displaystyle\sum_{i=1}^6f(r_i)\frac{\frac{3\pi}2-0}6=\pi\sum_{i=1}^6\sin\frac{i\pi}4

=\pi\left(\dfrac1{\sqrt2}+1+\dfrac1{\sqrt2}+0-\dfrac1{\sqrt2}-1\right)=\dfrac\pi{\sqrt2}\approx\boxed{2.221441}

Compare to the actual value of the integral, which is exactly 4.

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Maggie graphed the image of a 90 counterclockwise rotation about vertex A of . Coordinates B and C of are (2, 6) and (4, 3) and
Lemur [1.5K]

Answer:

A(2,2)

Step-by-step explanation:

Let the vertex A has coordinates (x_A,y_A)

Vectors AB and AB' are perpendicular, then

\overrightarrow {AB}=(2-x_A,6-y_A)\\ \\\overrightarrow {AB'}=(-2-x_A,2-y_A)\\ \\\overrightarrow {AB}\perp\overrightarrow {AB'}\Rightarrow \overrightarrow {AB}\cdot \overrightarrow {AB'}=0\Rightarrow (2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0

Vectors AC and AC' are perpendicular, then

\overrightarrow {AC}=(4-x_A,3-y_A)\\ \\\overrightarrow {AC'}=(1-x_A,4-y_A)\\ \\\overrightarrow {AC}\perp\overrightarrow {AC'}\Rightarrow \overrightarrow {AC}\cdot \overrightarrow {AC'}=0\Rightarrow (4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0

Now, solve the system of two equations:

\left\{\begin{array}{l}(2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0\\ \\(4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0\end{array}\right.\\ \\\left\{\begin{array}{l}-4-2x_A+2x_A+x_A^2+12-6y_A-2y_A+y^2_A=0\\ \\4-4x_A-x_A+x_A^2+12-3y_A-4y_A+y_A^2=0\end{array}\right.\\ \\\left\{\begin{array}{l}x_A^2+y_A^2-8y_A+8=0\\ \\x_A^2+y_A^2-5x_A-7y_A+16=0\end{array}\right.

Subtract these two equations:

5x_A-y_A-8=0\Rightarrow y_A=5x_A-8

Substitute it into the first equation:

x_A^2+(5x_A-8)^2-8(5x_A-8)+8=0\\ \\x_A^2+25x_A^2-80x_A+64-40x_A+64+8=0\\ \\26x_A^2-120x_A+136=0\\ \\13x_A^2-60x_A+68=0\\ \\D=(-60)^2-4\cdot 13\cdot 68=3600-3536=64\\ \\x_{A_{1,2}}=\dfrac{60\pm8}{2\cdot 13}=\dfrac{34}{13},2

Then

y_{A_{1,2}}=5\cdot \dfrac{34}{13}-8 \text{ or } 5\cdot 2-8\\ \\=\dfrac{66}{13}\text{ or } 2

Rotation by 90° counterclockwise about A(2,2) gives image points B' and C' (see attached diagram)

8 0
3 years ago
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Answer:

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Step-by-step explanation:

3.14 \times 10 ^{3}

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Answer:

x=75

Step-by-step explanation:

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