Get the answers you need at Westonci.ca, where our expert community is dedicated to providing you with accurate information. Discover detailed solutions to your questions from a wide network of experts on our comprehensive Q&A platform. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
Sagot :
Answer:
[tex]\displaystyle \int\limits^0_{\pi} {\sin (2x)} \, dx = 0[/tex]
General Formulas and Concepts:
Calculus
Integration
- Integrals
Integration Rule [Fundamental Theorem of Calculus 1]: [tex]\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)[/tex]
Integration Property [Multiplied Constant]: [tex]\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx[/tex]
U-Substitution
Step-by-step explanation:
Step 1: Define
Identify
[tex]\displaystyle \int\limits^0_{\pi} {\sin (2x)} \, dx[/tex]
Step 2: Integrate Pt. 1
Identify variables for u-substitution.
- Set u: [tex]\displaystyle u = 2x[/tex]
- [u] Differentiate: [tex]\displaystyle du = 2 \ dx[/tex]
- [Bounds] Switch: [tex]\displaystyle \left \{ {{x = 0 ,\ u = 2(0) = 0} \atop {x = \pi ,\ u = 2 \pi}} \right.[/tex]
Step 3: Integrate Pt. 2
- [Integral] Rewrite [Integration Property - Multiplied Constant]: [tex]\displaystyle \int\limits^0_{\pi} {\sin (2x)} \, dx = \frac{1}{2} \int\limits^0_{\pi} {2 \sin (2x)} \, dx[/tex]
- [Integral] U-Substitution: [tex]\displaystyle \int\limits^0_{\pi} {\sin (2x)} \, dx = \frac{1}{2} \int\limits^0_{2 \pi} {\sin u} \, du[/tex]
- Trigonometric Integration: [tex]\displaystyle \int\limits^0_{\pi} {\sin (2x)} \, dx = \frac{1}{2}(-\cos u) \bigg| \limits^0_{2 \pi}[/tex]
- Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: [tex]\displaystyle \int\limits^0_{\pi} {\sin (2x)} \, dx = \frac{1}{2}(0)[/tex]
- Simplify: [tex]\displaystyle \int\limits^0_{\pi} {\sin (2x)} \, dx = 0[/tex]
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Integration
Thank you for your visit. We are dedicated to helping you find the information you need, whenever you need it. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.