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elena-14-01-66 [18.8K]
3 years ago
10

Evaluate the integral. 3 2 t3i t t − 2 j t sin(πt)k dt

Mathematics
1 answer:
sveta [45]3 years ago
3 0

∫(t = 2 to 3) t^3 dt

= (1/4)t^4 {for t = 2 to 3}

= 65/4.

----

∫(t = 2 to 3) t √(t - 2) dt

= ∫(u = 0 to 1) (u + 2) √u du, letting u = t - 2

= ∫(u = 0 to 1) (u^(3/2) + 2u^(1/2)) du

= [(2/5) u^(5/2) + (4/3) u^(3/2)] {for u = 0 to 1}

= 26/15.

----

For the k-entry, use integration by parts with

u = t, dv = sin(πt) dt

du = 1 dt, v = (-1/π) cos(πt).


So, ∫(t = 2 to 3) t sin(πt) dt

= (-1/π) t cos(πt) {for t = 2 to 3} - ∫(t = 2 to 3) (-1/π) cos(πt) dt

= (-1/π) (3 * -1 - 2 * 1) + [(1/π^2) sin(πt) {for t = 2 to 3}]

= 5/π + 0

= 5/π.

Therefore,

∫(t = 2 to 3) <t^3, t√(t - 2), t sin(πt)> dt = <65/4, 26/15, 5/π>.

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Answer:

The length of q is 7.6 units

The measure of angle R is 31.9°

The measure of angle P is 66.1°

Step-by-step explanation:

Let us use the sine and cosine rules to solve the triangle

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∵ q² = p² + r² - 2(p)(r) cos(∠Q)

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∴ q² = 49 + 16 - 56 cos(82°)

∴ q² = 57.2063

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∴ q = 7.5635

- Round it to the nearest tenth

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∵ \frac{q}{sin(Q)}=\frac{r}{sin(R)}=\frac{p}{sin(P)}

- Let us find sin(R)

∴  \frac{7.5635}{sin(82)}=\frac{4}{sin(R)}

- By using cross multiplication

∴ 7.5635 × sin(R) = 4 × sin(82)

∴ 7.5635 sin(R) = 4 sin(82)

- Divide both sides 7.5635

∴ sin(R) = 0.5285

- Use sin^{-1} to find m∠R

∵ m∠R = sin^{-1} (0.5285)

∴ m∠R = 31.9042

- Round it to the nearest tenth

∴ The measure of angle R is 31.9°

∵ The sum of the measures of the interior angles of a Δ is 180°

∴ m∠P + m∠Q + m∠R = 180°

∴ m∠P + 82 + 31.9 = 180

∴ m∠P + 113.9 = 180

- Subtract 113.9 from both sides

∴ m∠P = 66.1°

∴ The measure of angle P is 66.1°

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Answer:

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