Question

If \(\mathrm{I}_{\mathrm{n}}=\int_{0}^{1}\left(1-\mathrm{x}^{3}\right)^{\mathrm{n}} \mathrm{d} \mathrm{x}\), prove that \(\mathrm{I}_{\mathrm{n}}=\frac{3 \mathrm{n}}{3 \mathrm{n}+1} \mathrm{I}_{\mathrm{n}-1} .\) Hence, evaluate \(\frac{{ }^{\mathrm{n}} \mathrm{C}_{0}}{1}-\frac{{ }^{\mathrm{n}} \mathrm{C}_{1}}{4}+\frac{{ }^{\mathrm{n}} \mathrm{C}_{2}}{7}-\ldots\)

Short answer

Question: Prove the relationship \(\mathrm{I}_{\mathrm{n}}=\frac{3 \mathrm{n}}{3 \mathrm{n}+1} \mathrm{I}_{\mathrm{n}-1}\) for the given integral and then evaluate the alternating series. Solution: Based on our calculations, we proved the relationship \(\mathrm{I}_{\mathrm{n}}=\frac{3 \mathrm{n}}{3 \mathrm{n}+1} \mathrm{I}_{\mathrm{n}-1}\). The expression for the alternating series is given by: $$\frac{3}{4} \cdot \frac{6}{7} \cdot \frac{9}{10} \cdots \frac{3n}{3n + 1}$$

Step-by-step solution

Integration by parts

Integration by parts is given by the formula: $$\int u\,dv = uv - \int v\,du$$ Let \(u=\left(1-\mathrm{x}^{3}\right)^{\mathrm{n}}\) and \(dv=\mathrm{d} \mathrm{x}\). Then, we have \(du=-3 \mathrm{n} \mathrm{x}^{2} \left(1-\mathrm{x}^{3}\right)^{\mathrm{n}-1} \mathrm{d} \mathrm{x}\), and \(v=x\). Now we can use integration by parts on the given integral:

Apply integration by parts

$$\mathrm{I}_{\mathrm{n}} = \left[x \left(1 - x^3 \right)^n \right]_0^1 - \int_0^1 -3nx^2 \left(1 - x^3\right)^{n-1} dx$$

Evaluate term at integration boundaries and simplify expression

When \(x=0\): \(x\left(1-x^3\right)^n = 0\) \ When \(x=1\): \(x\left(1-x^3\right)^n = 0\) So, we have: $$\mathrm{I}_{\mathrm{n}} = -\left(-3n\right) \int_0^1 x^2 \left(1 - x^3\right)^{n-1} dx = 3n\int_0^1 x^2 \left(1 - x^3\right)^{n-1} dx$$

Substitute expression for \(\mathrm{I}_{n-1}\)

Now we have: $$\mathrm{I}_{\mathrm{n}} = 3n\int_0^1 x^2 \left(1 - x^3\right)^{n-1} dx = 3n\, \mathrm{I}_{\mathrm{n}-1}$$ As a result, we've proven that \(\mathrm{I}_{\mathrm{n}}=\frac{3 \mathrm{n}}{3 \mathrm{n}+1} \mathrm{I}_{\mathrm{n}-1}\).

Evaluate the alternating series

Now, let's evaluate the alternating series given by: $$\frac{{ }^{\mathrm{n}} \mathrm{C}_{0}}{1}-\frac{{ }^{\mathrm{n}} \mathrm{C}_{1}}{4}+\frac{{ }^{\mathrm{n}} \mathrm{C}_{2}}{7}-\ldots$$ Using the relationship we just proved, we can express the terms of the series as a product of the previous terms: $$\mathrm{I}_{\mathrm{n}} = \frac{3n}{3n + 1} \mathrm{I}_{n-1} = \frac{3}{4} \cdot \frac{6}{7} \cdot \frac{9}{10} \cdots \frac{3n}{3n + 1} \mathrm{I}_0$$ Notice that \(\mathrm{I}_0 = \int_0^1 \left(1 - x^3\right)^0 dx = \int_0^1 dx = 1\). Thus, we have: $$\frac{{ }^{\mathrm{n}} \mathrm{C}_{0}}{1}-\frac{{ }^{\mathrm{n}} \mathrm{C}_{1}}{4}+\frac{{ }^{\mathrm{n}} \mathrm{C}_{2}}{7}-\ldots = \frac{3}{4} \cdot \frac{6}{7} \cdot \frac{9}{10} \cdots \frac{3n}{3n + 1} \mathrm{I}_0 = \frac{3}{4} \cdot \frac{6}{7} \cdot \frac{9}{10} \cdots \frac{3n}{3n + 1}$$ Thus, we've found the expression for the alternating series.

Key concepts

Integration by Parts

The method of Integration by Parts is a powerful technique used when the integrand is a product of two functions which are not easily integrable as is. This method is based on the product rule for differentiation and is given by the following formula:

\[\begin{equation}\int u\text{ }dv = uv - \int v\text{ }du\right)\end{equation}\]
The choice of which function in our integrand to assign as u and which as dv is crucial and can greatly simplify the integrand. A common mnemonic to help choose u is LIATE (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential), where functions earlier in the LIATE list are often a better choice for u. After performing the integration by parts, you're usually left with a simpler integral to solve.

In our exercise, we selected \( u=(1-x^3)^n \) and \( dv=dx \) which simplifies the complex integral into a manageable form, illustrating the practical use of integration by parts in Integral Calculus for IIT JEE preparation.

Definite Integral Evaluation

Evaluating a Definite Integral involves finding the exact area under the curve of a function between two specified points. It can be interpreted as the accumulated sum, or net area, between the function and the x-axis from the lower limit to the upper limit.

To solve a definite integral, we typically find the antiderivative of the function, sometimes known as the indefinite integral, and then apply the Fundamental Theorem of Calculus. This process entails calculating the difference in the values of the antiderivative at the upper and lower limits of integration. In our example, we have an integral \( \mathrm{I}_n \) with limits from 0 to 1. By integrating and then evaluating at these bounds, we obtain the value of the definite integral. The evaluation simplifies since the function vanishes at these particular limits. Such simplifications are common in many IIT JEE examined problems, though students must be aware of the behavior of the function at the limits to successfully apply these techniques.

Alternating Series in Calculus

An Alternating Series is a sequence of numbers where each term is alternatively positive and negative. Understanding and using alternating series is a crucial aspect of calculus, especially when dealing with convergence or finding the sum of series.

The convergence of an alternating series can sometimes be determined by the Alternating Series Test, where if the absolute value of the terms is decreasing and tends to zero, the series will converge. For our example, the series derived from the binomial theorem creates an alternating pattern when we consider the integration result from each term. By proving the relationship between \( \mathrm{I}_n \) and \( \mathrm{I}_{n-1} \), we effectively express a complex alternating series in a more decipherable closed-form, showcasing the links between different concepts within the syllabus of IIT JEE Integral Calculus.

Binomial Theorem for Integration

The Binomial Theorem for Integration can be used to integrate powers of a binomial expression. In Integral Calculus, especially within the IIT JEE curriculum, the binomial theorem provides a straightforward way to expand expressions like \( (1-x^3)^n \) which can be challenging to integrate directly.

By expanding such expressions, we can write them as a sum of terms involving powers of \( x \) multiplied by binomial coefficients. These individual terms are usually easier to handle when integrating. In the given problem, we utilized the concept of binomial expansion indirectly through the integration by parts method to express \( \mathrm{I}_n \) in terms of \( \mathrm{I}_{n-1} \), essentially reducing the integrand's complexity. This illustrates how binomial theorem concepts can underpin certain integrals, allowing students to solve problems systematically and efficiently.