Answer:
[tex]\displaystyle \int\limits^2_0 {f(x)} \, dx = \frac{159}{10}[/tex]
General Formulas and Concepts:
Calculus
Integration
Integration Rule [Reverse Power Rule]: [tex]\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C[/tex]
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]
Integration Property [Addition/Subtraction]: [tex]\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx[/tex]
Integration Property [Splitting Integral]: [tex]\displaystyle \int\limits^c_a {f(x)} \, dx = \int\limits^b_a {f(x)} \, dx + \int\limits^c_b {f(x)} \, dx[/tex]
Step-by-step explanation:
Step 1: Define
Identify.
[tex]\displaystyle f(x) = \left \{ {{9x^9 ,\ 0 \leq x \leq 1} \atop {4x^3 ,\ 1 \leq x \leq 2}} \right.[/tex]
[tex]\displaystyle \int\limits^2_0 {f(x)} \, dx = \ ?[/tex]
Step 2: Integrate
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Integration
Answer:
Step-by-step explanation:
[tex]\int\limits^1_0 {9x^9} \, dx +\int\limits^2_1 {4x^3} \, dx[/tex]
=[tex]\frac{9x^{10}}{10} |_0^1 + x^4|_1^2[/tex]
[tex]=\frac{9}{10}+2^4 -1 = 15\frac{9}{10}[/tex]
