If A is the area in the first quadrant enclosed by the curve C : 2x² - y + 1 = 0, the tangent to C a

Q: Why can't I just integrate the parabola from 0 to 1 and subtract the area of the triangle?

That is a brilliant geometric intuition! However, the 'triangle' formed by the lines is not a simple triangle because the tangent line is not parallel to the x-axis. The region is bounded by three distinct curves. Calculus is the only way to handle the curvature of the parabola accurately.

Q: I keep getting a negative area. What am I doing wrong?

A negative area usually means you have swapped the upper and lower functions. Remember, the area must be positive. Always ensure that $(y_{upper} - y_{lower}) \ge 0$ within your interval. If you get a negative result, check your subtraction order!

Q: Is the point $(1,3)$ definitely on the parabola?

Always verify! Plug $x=1$ into $y=2x^2+1$. You get $y=2(1)^2+1 = 3$. Yes, it lies perfectly on the curve. Never assume; always verify the coordinates before you start differentiating.

Question

If $A$ is the area in the first quadrant enclosed by the curve $C : 2x^2 - y + 1 = 0$, the tangent to $C$ at the point $(1, 3)$ and the line $x + y = 1$, then the value of $60A$ is................

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the Cartesian plane. You have a beautiful, upward-opening parabola defined by $C : 2x^2 - y + 1 = 0$, or more simply, $y = 2x^2 + 1$.

This curve is our primary actor, sitting with its vertex at $(0, 1)$. We are interested in the space trapped between this parabola and two specific lines.

The first line is $x + y = 1$, which we can rewrite as $y = 1 - x$. This is a downward-sloping boundary that intersects our parabola at the vertex $(0, 1)$.

### The Calculus Scalpel: Finding the Tangent

Now, we introduce the second line. We are given that there is a tangent to the curve at the point $(1, 3)$.

To find this tangent, we calculate the derivative of our parabola $y = 2x^2 + 1$:
$$\frac{dy}{dx} = 4x$$

At $x = 1$, the slope is $4$. Using the point-slope form, $y - 3 = 4(x - 1)$, we derive the equation of our tangent:
$$y = 4x - 1$$

### The Intersection: The 'Switch' in the Floor

The region enclosed by the curve, the line $y = 1 - x$, and the tangent $y = 4x - 1$ requires careful partitioning. The 'floor' of this region is not a single line.

From $x = 0$ to the intersection point, the floor is the line $y = 1 - x$. After that point, the floor becomes the tangent line $y = 4x - 1$.

We find where these two lines meet by setting them equal:
$$1 - x = 4x - 1 \Rightarrow 5x = 2 \Rightarrow x = \frac{2}{5}$$

### The Integration: The Heavy Lifting

We calculate the area in two phases. For the first phase, $A_1$, we integrate from $0$ to $2/5$:
$$A_1 = \int_{0}^{2/5} [(2x^2 + 1) - (1 - x)] \, dx = \int_{0}^{2/5} (2x^2 + x) \, dx$$

Integrating this, we evaluate:
$$\left[ \frac{2x^3}{3} + \frac{x^2}{2} ight]_0^{2/5} = \frac{2}{3} \left( \frac{8}{125} ight) + \frac{1}{2} \left( \frac{4}{25} ight) = \frac{16}{375} + \frac{30}{375} = \frac{46}{375}$$

For the second phase, $A_2$, we integrate from $2/5$ to $1$:
$$A_2 = \int_{2/5}^{1} [(2x^2 + 1) - (4x - 1)] \, dx = \int_{2/5}^{1} (2x^2 - 4x + 2) \, dx$$

Noting that $2x^2 - 4x + 2 = 2(x - 1)^2$, the integral becomes:
$$2 \left[ \frac{(x - 1)^3}{3} ight]_{2/5}^{1} = 0 - 2 \left( \frac{(-3/5)^3}{3} ight) = \frac{2}{3} \cdot \frac{27}{125} = \frac{54}{375}$$

### The Grand Finale

We sum our two areas to find the total area $A$:
$$A = \frac{46}{375} + \frac{54}{375} = \frac{100}{375} = \frac{4}{15}$$

The problem asks for the value of $60A$. Performing the final calculation:
$$60 	imes \left( \frac{4}{15} ight) = 4 	imes 4 = 16$$

The final result is **16**.

Animated Solution for Mathematics - Definite Integration: If $A$ is the area in the first quadrant enclosed by the curve $C : 2 x^{2} - y + 1 = 0$ , the tangent to $C$ at the point $(1, 3)$ and the line $x + y = 1$ , then the value of $60 A$ is................

Enter Numerical Value:

Visualized Solution

Identify the Curve $C$

Plot the Line $x + y = 1$

Find the Slope of Tangent at $(1, 3)$

Equation of the Tangent

Intersection of Tangent and Line

Visualizing the Enclosed Region

Setup Integral for Part 1: $A_{1}$

Calculate Area $A_{1}$

Setup Integral for Part 2: $A_{2}$

Calculate Area $A_{2}$

Total Area $A$

Final Value of $60 A$

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

The Calculus Scalpel

The Intersection

The Integration

The Grand Finale

Similar Questions

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Let $A$ be the area of the region ${(x, y) : y \geq x^{2}, y \geq (1 - x)^{2}, y \leq 2 x (1 - x)}$ . Then $540 A$ is equal to ______.

Enter Numerical Value:

Visualized Solution

Identify the Curve .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Plot the Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x+y=1

Find the Slope of Tangent at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (1,3)

Equation of the Tangent

Intersection of Tangent and Line

Visualizing the Enclosed Region

Setup Integral for Part 1: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A1​

Calculate Area .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A1​

Setup Integral for Part 2: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A2​

Calculate Area .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A2​

Total Area .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Final Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 60A