Mathematics, 31.01.2022 07:20 victor77

Evaluate the definite integral from pi/3 to pi/2 of (x+cosx) dx

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$\displaystyle \int\limits^{\frac{\pi}{2}}_{\frac{\pi}{3}} {(x + \cos x)} \, dx = \frac{5 \pi ^2}{72} + 1 - \frac{\sqrt{3}}{2}$

General Formulas and Concepts:

Calculus

Integration

Integrals

Integration Rule [Reverse Power Rule]: $\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C$

Integration Rule [Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)$

Integration Property [Addition/Subtraction]: $\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx$

Step-by-step explanation:

Step 1: Define

Identify.

$\displaystyle \int\limits^{\frac{\pi}{2}}_{\frac{\pi}{3}} {(x + \cos x)} \, dx$

Step 2: Integrate

[Integral] Rewrite [Integration Property - Addition/Subtraction]: $\displaystyle \int\limits^{\frac{\pi}{2}}_{\frac{\pi}{3}} {(x + \cos x)} \, dx = \int\limits^{\frac{\pi}{2}}_{\frac{\pi}{3}} {x} \, dx + \int\limits^{\frac{\pi}{2}}_{\frac{\pi}{3}} {\cos x} \, dx$ [Left Integral] Integration Rule [Reverse Power Rule]: $\displaystyle \int\limits^{\frac{\pi}{2}}_{\frac{\pi}{3}} {(x + \cos x)} \, dx = \frac{x^2}{2} \bigg| \limits^{\frac{\pi}{2}}_{\frac{\pi}{3}} + \int\limits^{\frac{\pi}{2}}_{\frac{\pi}{3}} {\cos x} \, dx$ [Right Integral] Trigonometric Integration: $\displaystyle \int\limits^{\frac{\pi}{2}}_{\frac{\pi}{3}} {(x + \cos x)} \, dx = \frac{x^2}{2} \bigg| \limits^{\frac{\pi}{2}}_{\frac{\pi}{3}} + \sin x \bigg| \limits^{\frac{\pi}{2}}_{\frac{\pi}{3}}$ Integration Rule [Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^{\frac{\pi}{2}}_{\frac{\pi}{3}} {(x + \cos x)} \, dx = \frac{5 \pi ^2}{72} + \bigg( 1 - \frac{\sqrt{3}}{2} \bigg)$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration

Answer from: tynasiaparks13

0.8194

Step-by-step explanation:

$\int\limits^\frac{\pi}{2} _\frac{\pi}{3} {x+cos(x)} \, dx\\\\=\frac{1}{2}x^2+sin(x)\Bigr|_{\frac{\pi}{3}}^{\frac{\pi}{2}}\\\\=[\frac{1}{2}(\frac{\pi}{2})^2+sin(\frac{\pi}{2})]-[\frac{1}{2}(\frac{\pi}{3})^2+sin(\frac{\pi}{3})]\\\\=[\frac{1}{2}(\frac{\pi^2}{4})+1]-[\frac{1}{2}(\frac{\pi^2}{9})+\frac{\sqrt{3}}{2}]\\ \\ =\frac{\pi^2}{8}-\frac{\pi^2}{18}+1-\frac{\sqrt{3}}{2}\\ \\ \approx0.8194$

Answer from: Quest

The answer would be b ! explanation: the formula for sa *surface area * is 2×l×w+2×l×h+2×w×h so in this case l=4w=12h=8so you use the formula to solve it and it give you the answer of 352^cm^2
Someone me. it would be nice if u could also explain how u ot the answer and how u do the equation

Someone me. it would be nice if u could also explain how u ot the answer and how u do the equation

Answer from: Quest

the value of x is 9 ⇒ answer b

step-by-step explanation:

* we can solve this problem using cosine and sine rule

- in δ abc

∵ ab = 36 , bc = 28 , ac = 16

- lets find the measure of angle b using the cosine rule

∵ cos b = [ab² + bc² - ac²]/2(ab)(bc)

∴ cos b = [36² + 28² - 16²]/2(36)(28)

∴ cos b = 19/21

∴ m∠b = cos^-1(19/21) ≅ 25°

- lets find the measure of angle a using the sine rule

∵ sina/bc = sinb/ac = sinc/ab

∴ sina/28 = sin25°/16 ⇒ by using cross multiplication

∴ sina = 28 × sin25°/16

∴ sina = 0.7396

∴ m∠a = sin^-1(0.7396) ≅ 48°

- from the figure m∠abd = m∠dbc

∵ m∠abc = 25°

∴ m∠abd = 25°/2 = 12.5°

- in δabd

∵ m∠a = 48° , m∠abd = 12.5°

∵ the sum of the measures of the interior angles of any δ is 180°

∴ m∠adb = 180° - (48° + 12.5°) = 119.5°

∵ ab = 36

∵ ad = x

- use the sine rule to find x

∵ sin∠adb/ab = sin∠abd/x

∴ sin119.5°/36 = sin12.5°/x ⇒ by using cross multiplication

∴ x = 36 × sin12.5°/sin119.5° = 8.95 ≅ 9

* the value of x is 9

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Evaluate the definite integral from pi/3 to pi/2 of (x+cosx) dx
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