Question

Evaluate by using the Fundamental Theorem of Calculus. The

integral(1 + 3y - y2)dy from 0 to 4. Round to tenths.

Answer 1

`\displaystyle \int\limits^4_0 {(1 + 3y - y^2)} \, dy = (20)/(3)`

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

1. Brackets
2. Parenthesis
3. Exponents
4. Multiplication
5. Division
6. Addition
7. Subtraction

• Left to Right

Calculus

Integrals

• Definite Integrals

Integration Constant C

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 [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`

Explanation:

Step 1: Define

Identify

`\displaystyle \int\limits^4_0 {(1 + 3y - y^2)} \, dy`

Step 2: Integrate

1. [Integral] Rewrite [Integration Property - Addition/Subtraction]:
   `\displaystyle \int\limits^4_0 {(1 + 3y - y^2)} \, dy = \int\limits^4_0 {} \, dy + \int\limits^4_0 {3y} \, dy - \int\limits^4_0 {y^2} \, dy`
2. [2nd Integral] Rewrite [Integration Property - Multiplied Constant]:
   `\displaystyle \int\limits^4_0 {(1 + 3y - y^2)} \, dy = \int\limits^4_0 {} \, dy + 3\int\limits^4_0 {y} \, dy - \int\limits^4_0 {y^2} \, dy`
3. [Integrals] Reverse Power Rule:
   `\displaystyle \int\limits^4_0 {(1 + 3y - y^2)} \, dy = y \bigg| \limits^4_0 + 3((y^2)/(2)) \bigg| \limits^4_0 - ((y^3)/(3)) \bigg| \limits^4_0`
4. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:
   `\displaystyle \int\limits^4_0 {(1 + 3y - y^2)} \, dy = 4 + 3(8) - (64)/(3)`
5. Simplify:
   `\displaystyle \int\limits^4_0 {(1 + 3y - y^2)} \, dy = (20)/(3)`

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

Unit: Integration

Book: College Calculus 10e