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Question

# Find the area of the region by subtracting the area of a triangular region from the area of a larger region. The region on or above the x-axis bounded by the curves y² = x + 3 and y = 2x.

Solution

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The area of the region which we need to evaluate is shaded by pink color. Firstly we will find the area of the region which is represented in the picture as sum of the pink area and the area of the purple triangle. So, from the picture we can notice that the line $x=1$ is upper function of the $x=y^2 -3$ on the interval $[0,2]$, in other words $1\geq y^3 -3$ for each $x \in [0,2]$. Therefore the sum-area we can obtain by

\begin{align*} \text{Area}&=\int\limits_{0}^{2} \left[1-\left(y^2 -3\right)\right]\dd{y}=\int\limits_{0}^{2} \left(4-y^2\right)\dd{y}\\\ &=\left[4y-\dfrac{y^3}{3}\right] \bigg|_{0}^{2}=4\cdot 2-\dfrac{8}{3}=\dfrac{16}{3} \end{align*}

Note that the upper limit of integration for this area we got by equalizing $x=y^2 -3$ with $x=1$.

Finally, the pink shaded area we will get by subtracting the area of the purple triangle from previously obtained area. We can notice that the purple triangle is the right-angled triangle, so its area is given as product of cathetus, i.e.

\begin{align*} \text{Purple triangle}=\dfrac{1 \cdot 2}{2}=1 \end{align*}

Thus,

\begin{align*} \text{Pink area}&=\text{Area} -\text{Purple triangle}=\dfrac{16}{3}-1=\dfrac{13}{3} \end{align*}

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