# Find the area bounded by the graphs of the indicated equations over the given interval (when...

## Question:

Find the area bounded by the graphs of the indicated equations over the given interval (when stated). Compute an answer to three decimals, if necessary.

{eq}y = 5e^x - 3; y = 0; 0 \leq x \leq 2 {/eq}

## Finding the Area:

Let us find the area for the given equations through integration by using the given limit {eq}0 \leq x \leq 2. {/eq}

The general form of area is,

{eq}Area = \int_{a}^{b} \left[ f \left(x \right) - g \left(x \right) \right] \ dx {/eq}

We have to consider that {eq}y = f \left(x \right) \ and \ g \left(x \right) {/eq} and then apply in the general form to get a solution.

## Answer and Explanation:

The given equations are:

{eq}y = 5e^x - 3 \\ y = 0 {/eq}

Let us consider the given equations into:

{eq}f \left(x \right) = 5e^x - 3 \\ g \left(x \right) = 0 {/eq}

The given limit is:

{eq}0 \leq x \leq 2 {/eq}

Now we are going to find the area of the given equations:

{eq}\begin{align*} Area &= \int_{a}^{b} \left[ f \left(x \right) - g \left(x \right) \right] \ dx \\ &= \int_{0}^{2} \left[5e^x - 3 - 0 \right] \ dx \\ &= \int_{0}^{2} \left[5e^x - 3 \right] \ dx \\ &= \left[ 5e^x - 3x \right] _{0}^{2} \\ &= \left[5e^2 - 3 \cdot 2 - \left( 5e^0 - 0 \right) \right] \\ &= \left[5e^2 - 6 - 5 \cdot 1 \right] \\ &= \left[5e^2 - 6 - 5 \right] \\ &= \left[5e^2 - 11 \right] \\ \therefore Area &= 25.94528 \end{align*} {/eq}

Therefore, the area value is {eq}\displaystyle 25.94528 {/eq}.

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from 6th-8th Grade Math: Practice & Review

Chapter 33 / Lesson 3