Question

Need to evaluate integral pls help me

Answer 1

Answer:

Step-by-step explanation:

$\int\limits^1_0 {9x^9} \, dx +\int\limits^2_1 {4x^3} \, dx$

=$\frac{9x^{10}}{10} |_0^1 + x^4|_1^2$

$=\frac{9}{10}+2^4 -1 = 15\frac{9}{10}$

Answer 2

Answer:

$\displaystyle \int\limits^2_0 {f(x)} \, dx = \frac{159}{10}$

General Formulas and Concepts:

Calculus

Integration

• Integrals
• Integral Notation

Integration Rule [Reverse Power Rule]:                                                               $\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C$

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

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

Integration Property [Addition/Subtraction]:                                                       $\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx$

Integration Property [Splitting Integral]:                                                               $\displaystyle \int\limits^c_a {f(x)} \, dx = \int\limits^b_a {f(x)} \, dx + \int\limits^c_b {f(x)} \, dx$

Step-by-step explanation:

Step 1: Define

Identify.

$\displaystyle f(x) = \left \{ {{9x^9 ,\ 0 \leq x \leq 1} \atop {4x^3 ,\ 1 \leq x \leq 2}} \right.$

$\displaystyle \int\limits^2_0 {f(x)} \, dx = \ ?$

Step 2: Integrate

1. [Integral] Rewrite [Integration Property - Splitting Integral]:                       $\displaystyle \int\limits^2_0 {f(x)} \, dx = \int\limits^1_0 {f(x)} \, dx + \int\limits^2_1 {f(x)} \, dx$
2. [Integrand] Substitute in function:                                                               $\displaystyle \int\limits^2_0 {f(x)} \, dx = \int\limits^1_0 {9x^9} \, dx + \int\limits^2_1 {4x^3} \, dx$
3. [Integrals] Rewrite [Integration Property - Multiplied Constant]:               $\displaystyle \int\limits^2_0 {f(x)} \, dx = 9 \int\limits^1_0 {x^9} \, dx + 4 \int\limits^2_1 {x^3} \, dx$
4. [Integrals] Integration Rule [Reverse Power Rule]:                                    $\displaystyle \int\limits^2_0 {f(x)} \, dx = 9 \bigg( \frac{x^{10}}{10} \bigg) \bigg| \limits^1_0 + 4 \bigg( \frac{x^4}{4} \bigg) \bigg| \limits^2_1$
5. Integration Rule [Fundamental Theorem of Calculus 1]:                            $\displaystyle \int\limits^2_0 {f(x)} \, dx = 9 \bigg( \frac{1}{10} \bigg) + 4 \bigg( \frac{15}{4} \bigg)$
6. Simplify:                                                                                                         $\displaystyle \int\limits^2_0 {f(x)} \, dx = \frac{159}{10}$

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

Unit: Integration