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Find the area bounded by y=x^2 - 4x and y= 3x+8.

Question:

Find the area bounded by {eq}\displaystyle y=x^{2}-4x \ and \ y=3x+8 {/eq}

Area Between Curves

The area between two curves {eq}f(x) {/eq} and {eq}g(x) {/eq} between {eq}x=a {/eq} and {eq}x=b {/eq} with {eq}f(x)\geq g(x) \text{ for all } x\in [a,b]. {/eq} can be found by evaluating

{eq}\displaystyle \int_a^b (f(x)-g(x))\, dx {/eq}

Answer and Explanation:

Below is the graph of {eq}y=x^2-4x {/eq} and {eq}y=3x+8. {/eq}

ENTER TITLE TEXT HERE

Note that {eq}y=3x+8 {/eq} and the curves intersect when

{eq}3x+8=x^2-4x\\ x...

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Learn more about this topic:

Approximating Definite Integrals on a Graphing Calculator
Approximating Definite Integrals on a Graphing Calculator

from Saxon Calculus Homeschool: Online Textbook Help

Chapter 7 / Lesson 8
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