Question

The area of the region under the curve of the function f(x)=5x+7 on the interval [1,b] is 88 square units, where b>1. What is the value of b.

Answer 1

Answer:5

Explanation:

got it right

Answer 2

`\displaystyle b = 5`

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 [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`

Area of a Region Formula:
`\displaystyle A = \int\limits^b_a {[f(x) - g(x)]} \, dx`

Explanation:

Step 1: Define

Identify

`\displaystyle \int\limits^b_1 {5x + 7} \, dx = 88`

Step 2: Solve

1. [Integral] Rewrite [Integration Property - Addition/Subtraction]:
   `\displaystyle \int\limits^b_1 {5x} \, dx + \int\limits^b_1 {7} \, dx = 88`
2. [Integrals] Rewrite [Integration Property - Multiplied Constant]:
   `\displaystyle 5 \int\limits^b_1 {x} \, dx + 7 \int\limits^b_1 {} \, dx = 88`
3. [Integrals] Integration Rule [Reverse Power Rule]:
   `\displaystyle 5 \bigg( (x^2)/(2) \bigg) \bigg| \limits^b_1 + 7(x) \bigg| \limits^b_1 = 88`
4. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:
   `\displaystyle 5 \bigg( (b^2)/(2) - (1)/(2) \bigg) + 7(b - 1) = 88`
5. Simplify:
   `\displaystyle (5b^2)/(2) - (5)/(2) + 7b - 7 = 88`
6. Isolate:
   `\displaystyle (5b^2)/(2) + 7b = (195)/(2)`
7. Solve:
   `\displaystyle b = (-39)/(5) ,\ 5`

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

Unit: Integration