Answer:
[tex]\displaystyle \int\limits^0_{\pi} {\sin (2x)} \, dx = 0[/tex]
General Formulas and Concepts:
Calculus
Integration
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
