The area (in sq. units) bounded by the parabola y = x² - 1, the tangent at the point (2, 3) to it an

Q: Why do we use $y_{upper} - y_{lower}$ in the integral?

The definite integral $\int_{a}^{b} f(x) dx$ calculates the signed area between the curve $f(x)$ and the x-axis. When you subtract the lower function from the upper function, you are essentially calculating the height of an infinitesimal vertical strip at any point $x$, which is $(y_{upper} - y_{lower})$. Integrating this height over the interval $[a, b]$ gives the total area of the region.

Q: How do I know which function is the upper boundary?

You can determine this by testing a point within the interval $[0, 2]$. For example, at $x=1$, the parabola $y = x^2 - 1$ gives $y = 0$, while the tangent line $y = 4x - 5$ gives $y = -1$. Since $0 > -1$, the parabola is the upper boundary.

Q: Can I use the formula for the area between a parabola and a chord?

Yes, you can! The area between a parabola and a chord is $\frac{1}{6} |a| (x_2 - x_1)^3$, where $a$ is the coefficient of $x^2$. However, this formula only applies when the region is bounded *only* by the parabola and the chord. In this problem, the y-axis is an additional boundary, so you must use the definite integral method to account for the specific limits.

Question

The area (in sq. units) bounded by the parabola $y = x^2 - 1$, the tangent at the point $(2, 3)$ to it and the y-axis is :

Options:

(A) $\frac{14}{3}$

(B) $\frac{56}{3}$

(C) $\frac{8}{3}$

(D) $\frac{32}{3}$

Sigma Percentile · Accepted Answer

### The Geometry of Constraints: A Journey into Area

Imagine you are standing on a coordinate plane, looking at the elegant curve of a parabola defined by $y = x^2 - 1$. It is a simple, symmetric shape, its vertex resting gracefully at $(0, -1)$.

Now, let us place a marker at the point $P(2, 3)$ on this curve. This point is not just a coordinate; it is the anchor for a line that will slice through the plane, creating a bounded region that we are tasked to measure.

This is the essence of calculus: taking a complex, enclosed space and breaking it down into manageable, solvable parts.

### The Tangent Hunt

Before we can calculate the area, we must define our boundaries. We know the parabola, but the tangent line at $P(2, 3)$ is a mystery we must solve.

In the language of calculus, the slope of a tangent line at any point is the derivative of the function at that point. We take our equation $y = x^2 - 1$ and apply the derivative operator:

$$
\frac{dy}{dx} = \frac{d}{dx}(x^2 - 1) = 2x
$$

At our point of interest, where $x = 2$, the slope $m$ becomes $m = 2(2) = 4$.

With a slope of $4$ and a point $(2, 3)$, we use the point-slope form $y - y_1 = m(x - x_1)$ to find the equation:

$$
y - 3 = 4(x - 2)
$$

Simplifying this, we get the equation of our tangent line: $y = 4x - 5$.

### Defining the Region

Now, look at the graph. We have three distinct boundaries: the parabola $y = x^2 - 1$ acting as the upper boundary, the tangent line $y = 4x - 5$ acting as the lower boundary, and the y-axis ($x = 0$) acting as the left wall.

The region is enclosed from $x = 0$ to $x = 2$. This is the space we need to measure.

It is a beautiful, curved triangle of sorts, and integration is the perfect tool to find its area.

### The Integral

To find the area $A$, we set up the definite integral:

$$
A = \int_{0}^{2} (y_{upper} - y_{lower}) \, dx
$$

Substituting our functions, we get:

$$
A = \int_{0}^{2} [(x^2 - 1) - (4x - 5)] \, dx
$$

Distributing the negative sign, the integrand becomes $x^2 - 1 - 4x + 5$, which simplifies to $x^2 - 4x + 4$.

Look closely at this expression. It is a perfect square: $(x - 2)^2$. The integral now looks much friendlier:

$$
A = \int_{0}^{2} (x - 2)^2 \, dx
$$

Using the power rule for integration, we find the antiderivative:

$$
\left[ \frac{(x - 2)^3}{3} \right]_{0}^{2}
$$

Plugging in the upper limit $2$, we get $0$. Plugging in the lower limit $0$, we get:

$$
\frac{(0 - 2)^3}{3} = \frac{-8}{3}
$$

Subtracting the lower limit from the upper limit, we arrive at:

$$
0 - \left( -\frac{8}{3} \right) = \frac{8}{3}
$$

### Conclusion

The area of this bounded region is exactly **$\frac{8}{3}$ square units**.

It is a testament to the power of calculus that we can take a seemingly abstract problem and find such a precise, elegant solution. Keep practicing, keep visualizing, and remember: every complex problem is just a series of simple steps waiting to be connected.

Animated Solution for Mathematics - Definite Integration: The area (in sq. units) bounded by the parabola $y = x^{2} - 1$ , the tangent at the point $(2, 3)$ to it and the y-axis is :

Select Answer:

Visualized Solution

Visualizing the Parabola and Point

Strategy for the Tangent Line

Differentiating the Curve

Calculating the Slope

Equation of the Tangent Line

Identifying the Bounded Region

Setting up the Area Integral

Substituting the Functions

Simplifying the Integrand

Integrating the Expression

Applying the Limits

Final Area Calculation

The Sigma Insight: Area Bounded by Curves

The Geometry of Constraints

The Tangent Hunt

Defining the Region

The Integral

Conclusion

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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; } The area (in sq. units) bounded by the parabola y=x2−1, the tangent at the point (2,3) to it and the y-axis is :