Question Statement

Find the area of the region bounded by the curve $y^2=4x$ and the line $x=3$.

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Background and Explanation

This problem requires calculating the area of a region bounded by a parabola and a vertical line using definite integration. The curve $y^2=4x$ is a right-opening parabola symmetric about the $x$-axis, so we can calculate the area of the upper half and double it to account for both the positive and negative $y$-values.

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Solution

Setting up the Integral

The given curve is $y^2=4x$, which represents a parabola opening to the right with its vertex at the origin. Solving for $y$:

$y=\pm 2\sqrt{x}$

This gives us two branches:

• Upper branch: $y=2\sqrt{x}$ (above the $x$-axis)
• Lower branch: $y=-2\sqrt{x}$ (below the $x$-axis)

To find where the curve intersects the $x$-axis (the starting point of our region), we set $y=0$:

$0=\pm 2\sqrt{x}⟹x=0$

The region extends from $x=0$ (the vertex) to $x=3$ (the given vertical line). Due to symmetry about the $x$-axis, we calculate the area of the upper half and multiply by 2:

$\text{Area}=2{\int }_0^3(2\sqrt{x})dx=4{\int }_0^3{x}^{1/2}dx$

Evaluating the Integral

Using the power rule for integration $\int x^{n}dx=\frac{x^{n+1}}{n+1}$:

$\begin{array}{rl}\text{Area}& =4{\int }_0^3{x}^{1/2}dx\\ & =4{\left[\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]}_0^3\\ & =4{\left[\frac{x^{3/2}}{3/2}\right]}_0^3\\ & =4\cdot \frac{2}{3}{\left[x^{3/2}\right]}_0^3\\ & =\frac{8}{3}\left[(3{)}^{3/2}-0\right]\end{array}$

Simplifying the Result

Simplify $(3{)}^{3/2}$:

$(3{)}^{3/2}=(3^3{)}^{1/2}=\sqrt{27}=\sqrt{9\times 3}=3\sqrt{3}$

Substituting back:

$\begin{array}{rl}\text{Area}& =\frac{8}{3}\cdot 3\sqrt{3}\\ & =8\sqrt{3}\text{ sq. units}\end{array}$

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Key Formulas or Methods Used

• Area between curves: $\text{Area}={\int }_a^b[y_{\text{upper}}-y_{\text{lower}}]dx$ (or using symmetry: $2\times \text{area of upper half}$)
• Power rule for integration: $\int x^{n}dx=\frac{x^{n+1}}{n+1}+C$ for $n\ne -1$
• Exponential simplification: $a^{m/n}=\sqrt[n]{a^m}=(\sqrt[n]{a}{)}^m$

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Summary of Steps

1. Identify the curve: Recognize $y^2=4x$ as a parabola and solve for $y$ to get $y=\pm 2\sqrt{x}$.
2. Determine bounds: Find that the curve meets the $x$-axis at $x=0$, so the region extends from $x=0$ to $x=3$.
3. Set up the integral: Use symmetry to write $\text{Area}=2{\int }_0^{3}2\sqrt{x}dx=4{\int }_0^3{x}^{1/2}dx$.
4. Integrate: Apply the power rule to get $\frac{8}{3}{x}^{3/2}$ evaluated from 0 to 3.
5. Evaluate and simplify: Calculate $\frac{8}{3}(3{)}^{3/2}=\frac{8}{3}\cdot 3\sqrt{3}=8\sqrt{3}$ square units.