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Find the area of the region bounded above by the graph of g(x) = 1 and below by the graph of f(x)...

Question:

Find the area of the region bounded above by the graph of {eq}g(x) = 1 {/eq} and below by the graph of {eq}f(x) = x^{2} + 2x - 2 {/eq}.

Areas as Single Integrals


The area of a region bounded above by {eq}\displaystyle y=g(x), {/eq} bellow, by {eq}\displaystyle y=f(x) , {/eq} between {eq}\displaystyle x=a, x=b, {/eq}

is given as {eq}\displaystyle \int_a^b \left[ g(x)-f(x)\right]\ dx. {/eq}

Sometimes, we need to determine the limits of integration, a and b by finding the points of intersection of the two cures.

Evaluating definite integrals, we may need to the following formula

{eq}\displaystyle \int x^{n}\ dx =\frac{1}{n+1}x^{n+1}+C, n\neq -1, C - \text{ constant}. {/eq}

Answer and Explanation:


To find the area of the region bounded above by the line {eq}\displaystyle g(x)=1, {/eq} and below by the parabola {eq}\displaystyle f(x)=x^2+2x-2 {/eq}

we will first describe the region.

The two curves intersect when {eq}\displaystyle f(x)=g(x)\iff x^2+2x-2=1\iff x^2+2x-3=0\implies x=-3 \text{ and } x=1. {/eq}

Therefore, the area of the region is given by the following definite integral

{eq}\displaystyle \begin{align} \text{Area}&=\int_{-3}^1\left[g(x)-f(x)\right]\ dx\\ &=\int_{-3}^1\left(-x^2-2x+3\right)\ dx\\ &=\left(-\frac{x^3}{3}-x^2+3x\right) \bigg\vert_{-3}^1\\ &=-\frac{1-27}{3}-1+9+3-9\\ &=\boxed{\frac{32}{3} }. \end{align} {/eq}


Learn more about this topic:

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Evaluating Definite Integrals Using the Fundamental Theorem

from AP Calculus AB: Exam Prep

Chapter 16 / Lesson 2
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