Find the area bounded by the curve x² = 4y and the straight line x = 4y - 2.

Q: Wait, why do we subtract the parabola from the line? How do I know which one is the upper curve?

That is a fantastic question! The rule is simple: in the region where the curves intersect, the 'upper' curve is the one with the greater $y$-value for any given $x$. If you are ever unsure, just pick a test point between your intersection limits (like $x=0$) and plug it into both equations. Here, at $x=0$, the line gives $y = 0.5$ and the parabola gives $y = 0$. Since $0.5 > 0$, the line is definitely on top!

Q: Can I integrate with respect to $y$ instead of $x$?

Yes, you absolutely can! If you integrate with respect to $y$, you would express $x$ as a function of $y$ for both curves. The parabola becomes $x = \pm 2\sqrt{y}$ and the line becomes $x = 4y - 2$. However, this is often more complex because the parabola has two branches. Integrating with respect to $x$ is usually the path of least resistance for this specific problem.

Q: What happens if I get a negative area? Did I do something wrong?

If you calculate a negative area, it usually means you swapped the upper and lower curves in your integral setup. Remember, area must always be positive. If you get a negative result, don't panic—it just means you calculated $\int (y_{lower} - y_{upper}) dx$ instead of $\int (y_{upper} - y_{lower}) dx$. Just take the absolute value, or better yet, re-check your upper and lower curve identification!

Question

Find the area bounded by the curve $x^2 = 4y$ and the straight line $x = 4y - 2$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a coordinate plane, looking at two distinct mathematical entities. First, we have the parabola $x^2 = 4y$, a graceful, upward-opening curve symmetric about the $y$-axis. Then, cutting across it like a sharp blade, we have the straight line $x = 4y - 2$.

Our goal is to find the area of the region trapped between these two. Before we touch a single integral, we must find the boundaries of this 'trap' by solving the equations simultaneously.

From the parabola, we know $4y = x^2$. Substituting this into the line equation $x = 4y - 2$, we get $x = x^2 - 2$.

Rearranging this, we arrive at the quadratic equation:

$$x^2 - x - 2 = 0$$

Factoring this, we find $(x - 2)(x + 1) = 0$, which gives us our limits of integration: $x = -1$ and $x = 2$. These are the vertical walls of our region.

### The Integral Setup

Now that we have our boundaries, we need to calculate the area. Think of the area as being composed of an infinite number of infinitesimally thin vertical strips.

The height of each strip is the difference between the upper curve and the lower curve. Looking at our graph, the line $y = \frac{x+2}{4}$ sits above the parabola $y = \frac{x^2}{4}$ between $x = -1$ and $x = 2$.

Thus, our integral is:

$$A = \int_{-1}^{2} \left( \frac{x+2}{4} - \frac{x^2}{4} \right) dx$$

To make our lives easier, we factor out the constant $\frac{1}{4}$, leaving us with:

$$A = \frac{1}{4} \int_{-1}^{2} (x + 2 - x^2) dx$$

### The Execution

Now, we apply the power rule of integration. The integral of $x$ is $\frac{x^2}{2}$, the integral of $2$ is $2x$, and the integral of $x^2$ is $\frac{x^3}{3}$.

Our expression becomes:

$$A = \frac{1}{4} \left[ \frac{x^2}{2} + 2x - \frac{x^3}{3} \right]_{-1}^{2}$$

Evaluating this at the upper limit $x = 2$:

$$A_{upper} = \frac{1}{4} \left[ \frac{4}{2} + 4 - \frac{8}{3} \right] = \frac{1}{4} \left[ 6 - \frac{8}{3} \right] = \frac{1}{4} \left[ \frac{10}{3} \right] = \frac{10}{12}$$

Now, for the lower limit $x = -1$:

$$A_{lower} = \frac{1}{4} \left[ \frac{1}{2} - 2 - \frac{-1}{3} \right] = \frac{1}{4} \left[ \frac{1}{2} - 2 + \frac{1}{3} \right] = \frac{1}{4} \left[ \frac{3-12+2}{6} \right] = \frac{1}{4} \left[ -\frac{7}{6} \right] = -\frac{7}{24}$$

Subtracting the lower limit from the upper limit, we get:

$$A = \frac{10}{12} - \left( -\frac{7}{24} \right) = \frac{20}{24} + \frac{7}{24} = \frac{27}{24}$$

Simplifying this fraction, we arrive at our final, elegant answer:

**$A = \frac{9}{8} = 1.125$ square units.**

Animated Solution for Mathematics - Definite Integration: Find the area bounded by the curve $x^{2} = 4 y$ and the straight line $x = 4 y - 2$ .

Enter Numerical Value:

Visualized Solution

Visualizing the Curves

Identifying the Objective

Finding Intersection Points

Algebraic Setup

Solving the Quadratic

Setting up the Integral

Identifying Upper and Lower Curves

Substituting into the Integral

Integrating the Expression

Evaluating at Upper Limit

Evaluating at Lower Limit

Final Calculation

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

The Integral Setup

The Execution

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Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Find the area bounded by the curve x2=4y and the straight line x=4y−2.