# Sketch the region bounded by the curves, and visually estimate the location of the centroid:

$\boldsymbol{ y \ = \ e^x, \ y \ = \ 0, \ x \ = \ 0, \ x \ = \ 5 }$

The aim of this question is to find the area under a bounded region with multiple constraints and to calculate the centroid of this bounded region.

To solve this question, we first find the area bounded by the region (say A). Then we calculate the x and y moments of the region (say $M_x$ & $M_y$). The moment is the measure of the tendency of a given region against rotation around the origin. Once we have these moments, we can calculate the centroid C using the following formula:

$C = \left( \dfrac{M_y}{A}, \dfrac{M_x}{A} \right)$

Step (1): The constraint of $y = 0$ is already fulfilled. To find the area bounded by the region $y \ = \ e^x$, we need to perform following integration:

$A = \int_{a}^{b} \bigg ( e^x \bigg ) dx$

Since the region is bounded by $x \ = \ 0$ and $x \ = \ 5$:

$A = \int_{0}^{5} \bigg ( e^x \bigg ) dx$

$\Rightarrow A = \bigg | e^x \bigg |_{0}^{5}$

$\Rightarrow A = e^{ (5) } \ – \ e^{ (0) }$

$\Rightarrow A = e^5 \ – \ 1$

Step (2): Calculating the $M_x$:

$M_x = \int_{0}^{5} \bigg ( e^x \bigg )^2 dx$

$\Rightarrow M_x = \bigg | \frac{ 1 }{ 2 } \bigg ( \frac{e^x}{2} \bigg ) (e^x) \bigg |_{0}^{5}$

$\Rightarrow M_x = \bigg | \frac{ e^{ 2x } }{ 4 } \bigg |_{0}^{5}$

$\Rightarrow M_x = \frac{ 1 }{ 4 } \bigg | e^{ 2x } \bigg |_{0}^{5}$

$\Rightarrow M_x = \frac{ 1 }{ 4 }\bigg ( e^{ 2(5) } – e^{ 2(0) } \bigg )$

$\Rightarrow M_x = \frac{ 1 }{ 4 }\bigg ( e^{ 2(5) } – 1 \bigg )$

Step (3): Calculating the $M_y$:

$M_x = \int_{0}^{5} \bigg ( xe^x \bigg ) dx$

$\Rightarrow M_y = \bigg | (x-1)e^x \bigg |_{0}^{5}$

$\Rightarrow M_y = \bigg ( (5-1)e^{(5)} -(0-1)e^{(0)} \bigg )$

$\Rightarrow M_y = 4e^5 + 1$

Step (4): Calculating the x-coordinate of centroid:

$C_x = \dfrac{M_x}{A}$

$C_x = \dfrac{ \dfrac{ 1 }{ 4 }\bigg ( e^{ 2(5) } – 1 \bigg )}{e^5-1}$

$C_x = \dfrac{ \dfrac{ 1 }{ 4 }\bigg ( (e^5)^2 – (1)^2 \bigg )}{e^5-1}$

$C_x = \dfrac{ \dfrac{ 1 }{ 4 }(e^5 – 1)(e^5 + 1) }{e^5-1}$

$C_x = \dfrac{ 1 }{ 4 }(e^5 + 1)$

$C_x = 37.35$

Step (5): Calculating the y-coordinate of centroid:

$C_y = \dfrac{M_y}{A}$

$C_y = \dfrac{4e^5 + 1}{e^5-1}$

$C_y = 4.0$

## Numerical Result

$Centroid \ = \ \left [ \ 37.35, \ 4.0 \ \right ]$

## Example

Given that $M_x = 30$, $M_y = 40$ and $A = 10$, find the coordinates of the centroid of the bounded region.

x-coordinate of centroid $C_x$ can be calculated using:

$C_x = \dfrac{M_x}{A} = \dfrac{30}{10} = 3$

y-coordinate of centroid $C_y$ can be calculated using:

$C_y = \dfrac{M_y}{A} = \dfrac{40}{10} = 4$

So:

$Centroid \ = \ \left [ \ 3, \ 4 \ \right ]$