Mathematics, 21.08.2019 16:30, lulustar13

Use integration by parts to integrate sin2x between pi and 0

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answered: leannamat2106

$\displaystyle \int\limits^0_{\pi} {\sin (2x)} \, dx = 0$

General Formulas and Concepts:

Calculus

Integration

Integrals

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

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

U-Substitution

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \int\limits^0_{\pi} {\sin (2x)} \, dx$

Step 2: Integrate Pt. 1

Identify variables for u-substitution.

Set u: $\displaystyle u = 2x$ [u] Differentiate: $\displaystyle du = 2 \ dx$ [Bounds] Switch: $\displaystyle \left \{ {{x = 0 ,\ u = 2(0) = 0} \atop {x = \pi ,\ u = 2 \pi}} \right.$

Step 3: Integrate Pt. 2

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^0_{\pi} {\sin (2x)} \, dx = \frac{1}{2} \int\limits^0_{\pi} {2 \sin (2x)} \, dx$ [Integral] U-Substitution: $\displaystyle \int\limits^0_{\pi} {\sin (2x)} \, dx = \frac{1}{2} \int\limits^0_{2 \pi} {\sin u} \, du$ Trigonometric Integration: $\displaystyle \int\limits^0_{\pi} {\sin (2x)} \, dx = \frac{1}{2}(-\cos u) \bigg| \limits^0_{2 \pi}$ Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^0_{\pi} {\sin (2x)} \, dx = \frac{1}{2}(0)$ Simplify: $\displaystyle \int\limits^0_{\pi} {\sin (2x)} \, dx = 0$

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

Unit: Integration

answered: Guest

the monthly balance will be $75,000

answered: Guest

hi there!

first we need to find out how much they are adding to each number to make the next number.

ex: 4 + 4 = 8, 8 + 8 = 16, and so on.

so continuing with 32,

32 + 32 = 64

64 + 64 = 128

128 + 128 = 256

256 + 256 = 512

4, 8, 16, 32, 64, 128, 256, 512

the 8th term of the sequence will be 512

hope this ! ~

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Use integration by parts to integrate sin2x between pi and 0...