Problem 103 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(\int f(x) d x=\Psi(x)\) then \(\int x^{5} f\left(x^{3}\right) d x\) is equal to (A) \(\frac{1}{3} x^{3} \Psi\left(x^{3}\right)-3 \int x^{3} \Psi\left(x^{3}\right) d x+C\) \([\mathbf{2 0 1 3}]\) (B) \(\frac{1}{3} x^{3} \Psi\left(x^{3}\right)-\int x^{2} \Psi\left(x^{3}\right) d x+C\) (C) \(\frac{1}{3}\left[x^{3} \Psi\left(x^{3}\right)-\int x^{3} \Psi\left(x^{3}\right) d x\right]+C\) (D) \(\frac{1}{3}\left[x^{3} \Psi\left(x^{3}\right)-\int x^{2} \Psi\left(x^{3}\right) d x\right]+C\)

Step-by-Step Solution

Answer

Option (D) is correct: \(\frac{1}{3}\left[x^{3} \Psi\left(x^{3}\right)-\int x^{2} \Psi\left(x^{3}\right) dx\right]+C\)."

Step 1: Identify the Integration Formula

To solve \(\int x^{5} f\left(x^{3}\right) dx\), we start by recognizing the formula of integration by parts \(\int u \, dv = uv - \int v \, du\). This problem will involve clever substitution and integration by parts.

Step 2: Choose Substitution

Let \(z = x^3\). Then \(dz = 3x^2 \, dx\). So, \(dx = \frac{dz}{3x^2}\) and \(x^{5} = \left(z^\frac{5}{3}\right)\). This will help simplify the expression to use integration by parts.

Step 3: Simplify the Integral

Substituting \(z = x^3\) into the integral gives \(\int x^5 f(x^3) \, dx = \int x^3 x^2 f(x^3) \, dx = \int \left(\frac{z}{3x^2}\right) \cdot z^{\frac{5}{3}} f(z) \, dz\).

Step 4: Apply Integration by Parts

To apply integration by parts, we use \(u = z^{\frac{5}{3}}\) and \(dv = f(z) \ \frac{dz}{3}\). Consequently, \(du = \frac{5}{3}z^{\frac{2}{3}} \, dz\), \(v = \Psi(z)\). Integrate by parts: \(\frac{1}{3}\left[z\Psi(z) - \int \frac{dz}{3}\cdot z \Psi(z)\right]\).

Step 5: Simplify and Match with Options

Simplifying gives \(\frac{1}{3}\left[x^3 \Psi(x^3) - \int x^2 \Psi(x^3) \, dx \right] + C\), which matches option D. Compare each step with the provided options to validate the correct solution.

Key Concepts

Substitution in Integration

Substitution in integration is a crucial technique in calculus, often used to simplify complex integrals. This technique involves changing variables to make the integration process more manageable.

It's somewhat like a change of perspective, where we temporarily replace a complex part of the function with a simpler variable.

• First, identify a part of the integral that you can isolate and substitute with a new variable. This component usually involves expressions that are, by themselves, quite complex or cumbersome to integrate directly.
• Replace this part with a chosen new variable. For example, in the exercise, the substitution was made by setting \( z = x^3 \).
• Substitute back and express all parts of the integral in terms of the new variable. This sometimes involves expressing \( dx \) in terms of the new variable's differential.

By making this substitution, the integral's form often becomes more straightforward, enabling easier application of integration techniques, such as integration by parts, as seen in the original exercise.
Integration by parts was essential after the substitution step, highlighting its importance in combining with other methods.

Definite and Indefinite Integrals

Understanding definite and indefinite integrals forms the backbone of integral calculus.
These integrals are used to find areas under curves and express antiderivatives, respectively, enabling us to solve a wide range of mathematical problems.

• An indefinite integral represents a family of functions and includes a constant of integration \( C \). It is often used to find the antiderivative of a function. For instance, when the solution labels a "+ C," it indicates an indefinite integral.
• A definite integral, on the other hand, computes the area under a curve between two points and does not require the constant \( C \). It provides a specific numerical value.

In the exercise, we're primarily dealing with indefinite integrals, where finding the antiderivative is the goal. Understanding these concepts is key to solving the problem efficiently. Recognizing when to use a definite versus an indefinite integral is crucial in calculus problem-solving.

Integral Calculus Methods

Integral calculus methods encompass several techniques to evaluate integrals, each with specific applications.
Mastering them requires practice and a solid understanding of the foundational concepts.

• Substitution, as discussed, simplifies the process of integrating complex functions by altering the variable of integration.
• Integration by Parts is another valuable technique, used when the integral is a product of two functions. It leverages the formula \( \int u \, dv = uv - \int v \, du \), which was used in the exercise after substitution.
• Remember, there are also methods like partial fraction decomposition and trigonometric integration for special classes of functions.

Each method has specific scenarios where it is most effective. Thus, being familiar with them allows for flexible problem-solving approaches.
In the exercise, a combination of substitution and integration by parts demonstrated how these techniques can work together to solve a problem efficiently.