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:

$\displaystyle b = 5$

General Formulas and Concepts:

Calculus

Integration

• Integrals

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]:                                                       $\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx$

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

Step-by-step 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]:                                     $\displaystyle 5 \bigg( \frac{x^2}{2} \bigg) \bigg| \limits^b_1 + 7(x) \bigg| \limits^b_1 = 88$
4. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:           $\displaystyle 5 \bigg( \frac{b^2}{2} - \frac{1}{2} \bigg) + 7(b - 1) = 88$
5. Simplify:                                                                                                         $\displaystyle \frac{5b^2}{2} - \frac{5}{2} + 7b - 7 = 88$
6. Isolate:                                                                                                           $\displaystyle \frac{5b^2}{2} + 7b = \frac{195}{2}$
7. Solve:                                                                                                             $\displaystyle b = \frac{-39}{5} ,\ 5$

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

Unit: Integration