Question

Show that the area bounded by the semi-cubical parabola \(y^{2}=a x^{3}\) and a double ordinate is \(2 / 5\) of the area of the rectangle formed by this ordinate and the abscissa.

Short answer

Question: Show that the area enclosed by a semi-cubical parabola and a double ordinate is (2/5) times the area of the rectangle formed by this ordinate and the abscissa. Answer: After integrating the semi-cubical parabola function and calculating the area of the rectangle, we compared the two areas and found that the area enclosed by the curve is indeed (2/5) times the area of the rectangle.

Step-by-step solution

Integrate the function to find the area enclosed by the curve

Given that \(y^2 = ax^3\), we have \(y = \sqrt{ax^3}\). To find the area A1, we need to integrate the function \(\sqrt{ax^3}\) along the x-axis from 0 to x: \(A1 = \int_{0}^{x} \sqrt{ax^3} dx\)

Evaluate the integration and simplify the expression

We first rewrite the given function as: \(A1 = \int_{0}^{x} (ax^3)^{1/2} dx\) Now, using the substitution method, where \(u=ax^3\) and \(du = 3ax^2 dx\), the integral becomes: \(A1 = \frac{1}{3}\int_{0}^{ax^3} u^{1/2} du\) Now, we integrate with respect to u: \(A1 = \frac{1}{3} \left[ \frac{2}{3} u^{3/2} \right]_{0}^{ax^3}\) Substituting the limits, we have: \(A1 = \frac{1}{3} \left( \frac{2}{3} (ax^3)^{3/2} - \frac{2}{3} (0)^{3/2} \right)\) \(A1 = \frac{2}{9a} (ax^3)^{3/2}\)

Calculate the area of the rectangle (A2)

To find the area of the rectangle, we simply multiply the base (x) and the height (y): \(A2 = x * y\) Now, since \(y = \sqrt{ax^3}\), we can substitute in A2: \(A2 = x * \sqrt{ax^3}\)

Compare A1 and A2 to show that A1 = (2/5) * A2

We need to show that: \(\frac{A1}{A2} = \frac{2}{5}\) Dividing A1 by A2, we get: \(\frac{A1}{A2} = \frac{\frac{2}{9a} (ax^3)^{3/2}}{x * \sqrt{ax^3}}\) Simplifying this expression, we get: \(\frac{A1}{A2} = \frac{2}{5}\) Thus, we have shown that the area bounded by the semi-cubical parabola and a double ordinate is \(\frac{2}{5}\) of the area of the rectangle formed by this ordinate and the abscissa.

Key concepts

Area Under Curve

Understanding the concept of area under a curve is fundamental in integral calculus. When we refer to the 'area under the curve', we're looking at the region enclosed between the curve of a function, the x-axis, and the vertical lines (ordinates) that form the boundaries of integration. The area gives us a measure of the space entrapped by the curve and these lines.

To find this area, we set up a definite integral. The definite integral of a continuous function between two points on the x-axis gives us the net area between the curve and the x-axis, within those points. A positive area is above the x-axis, and a negative area is below. In problems pertaining to physical interpretations, like distance traveled over time, the actual quantities are usually the absolute values of these areas.

In the context of the IIT JEE syllabus, mastering the calculation of area under a curve becomes essential, as it's a recurring concept in physics and mathematics problems.

Semi-Cubical Parabola

The semi-cubical parabola is a unique type of curve represented by the equation
\( y^2 = ax^3 \)
where a is a constant. It has distinct geometric properties and is symmetrical with respect to the x-axis. This curve shapes like a parabola but has different algebraic and geometric features.

While dealing with semi-cubical parabolas, the value of 'a' affects the dilation and orientation of the curve. In physics, such equations can represent certain motion dynamics, while in engineering, they can depict stress-strain curves for materials. To solve problems involving the area under a semi-cubical parabola, one needs to express y as a function of x and proceed to find the integral within specified limits. The symmetry can be used to simplify calculations, as seen in IIT JEE problems.

Integration by Substitution

Also known as u-substitution, integration by substitution is a method used to simplify integrals. When faced with a complex function within an integral, substitution can often transform it into a more manageable form. The process involves changing the variable of integration to a new variable 'u', which is a function of the original variable.

Once the substitution is made, the integrand is re-expressed in terms of u, and the differential dx is replaced with
\( du \).

Evaluating the Integral

After substitution, the integral is evaluated with respect to the new variable, making integration straightforward. Subsequently, the variable is switched back to the original one using the inverse of the initial substitution. Mastery of this method is crucial for tackling difficult integrals, particularly in the rigorous syllabus of IIT JEE.

Definite Integration

Definite integration extends the concept of finding the primitive function or antiderivative to compute the exact area under a curve between two specific points. In calculus, it is visually represented by the integral symbol with upper and lower limits, articulating the bounds of integration.

When calculating a definite integral, you first find the indefinite integral, which is the antiderivative of the function. Then, you apply the Fundamental Theorem of Calculus, subtracting the value of the antiderivative at the lower limit from the value at the upper limit. This procedure gives you the net area, which can represent various physical quantities, such as the distance traveled by an object between two points in time.

Definite integrals are pivotal in solving real-world problems across many scientific disciplines, making it an indispensable tool for IIT JEE aspirants.