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