Question

I really need this answr tonight please HELP!!! Use limits to find the area between the graph of the function and the x axis given by the definite integral. ∫_1^5▒(x^2-x+1)dx

Answer 1

Answer: 5\int_0^5 3^(3)\sqrt(x^(2)) dx

Explanation:

Answer 2

`\displaystyle \int\limits^5_1 {(x^2 - x + 1)} \, dx = (100)/(3)`

General Formulas and Concepts:

Calculus

Integration

• Integrals

Integration Rule [Reverse Power Rule]:
`\displaystyle \int {x^n} \, dx = (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 [Addition/Subtraction]:
`\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx`

Explanation:

Step 1: Define

Identify

`\displaystyle \int\limits^5_1 {(x^2 - x + 1)} \, dx`

Step 2: Integrate

1. [Integral] Rewrite [Integration Property - Addition/Subtraction]:
   `\displaystyle \int\limits^5_1 {(x^2 - x + 1)} \, dx = \int\limits^5_1 {x^2} \, dx - \int\limits^5_1 {x} \, dx + \int\limits^5_1 {} \, dx`
2. [Integrals] Integration Rule [Reverse Power Rule]:
   `\displaystyle \int\limits^5_1 {(x^2 - x + 1)} \, dx = \bigg( (x^3)/(3) - (x^2)/(2) + x \bigg) \bigg| \limits^5_1`
3. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:
   `\displaystyle \int\limits^5_1 {(x^2 - x + 1)} \, dx = (100)/(3)`

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

Unit: Integration