Question

If f is continuous and \(\int_{0} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}=4\), find \(\int_{0}^{3} x f\left(x^{2}\right) d x\)

Short answer

Answer: The value of the integral is 18.

Step-by-step solution

Perform substitution

Let's substitute \(u = x^2\). Now, differentiate with respect to x: $$ \frac{d u}{d x}=2 x d x $$ Now, let's solve for \(dx\): $$ d x=\frac{d u}{2 x} $$ Also, we have to change the limits of integration according to the substitution. When \(x=0\), \(u=0^2 = 0\) and when \(x=3\), \(u=3^2 = 9\).

Rewrite the integral using the substitution

Now, we rewrite the integral in terms of u: $$ \int_{0}^{3} x f\left(x^{2}\right) d x=\int_{0}^{9} \frac{1}{2} \sqrt{u} f(u) d u $$

Use the given information to evaluate the integral

Since we know that \(\int_{0}^{x} f(t) dt = 4\), we can identify that for any x within the range, the integral of f(t) will be 4. Now, we have to apply this knowledge to evaluate the integral with respect to u: $$ \int_{0}^{9} \frac{1}{2} \sqrt{u} f(u) d u = \frac{1}{2} \int_{0}^{9} \sqrt{u} f(u) d u = \frac{1}{2} \cdot \left(4\right) \cdot \left(9-0\right) $$

Calculate the final result

Now, multiplying the constants, we get the final value of the integral: $$ \frac{1}{2} \cdot 4 \cdot 9 = 2 \cdot 9 = 18 $$ Therefore, the value of \(\int_{0}^{3} xf\left(x^2\right)dx\) is 18.

Key concepts

U-Substitution

One of the most powerful techniques for evaluating integrals is \textbf{u-substitution}, also known as integration by substitution. This method is essentially the reverse process of the chain rule in differentiation. It is used when an integral contains a function and its derivative.
When approaching an integral that seems complex due to a composition of functions, we make a substitution to simplify the expression. We choose a new variable, typically denoted as 'u,' which is a function of 'x'. Then, we express 'dx' in terms of 'du' by differentiating 'u' with respect to 'x'.

Practical Steps in U-Substitution

• Identify a portion within the integral that can serve as 'u'.
• Differentiate 'u' to find 'du' in terms of 'dx'.
• Replace all the 'x' parts of the integral with the corresponding 'u' parts.
• Change the limits of integration if performing a definite integral.
• Integrate with respect to 'u'.
• Substitute back the original variables to find the solution if necessary.

In the step-by-step solution provided, the substitution was made by setting \(u = x^2\), which when differentiated led to \(du = 2xdx\), and simplified the integration process significantly.

Definite Integral Evaluation

Evaluating a \textbf{definite integral} involves calculating the integral of a function over a specific interval. The definite integral of a continuous function is a fundamental concept in integral calculus and can represent the accumulated sum, area under a curve, or the total net change.
To evaluate a definite integral, we find the antiderivative of the function, substitute the upper and lower limits of integration, and then find the difference. The notable point is that unlike indefinite integrals, definite integrals result in a specific numerical value.

Considerations for Definite Integrals

• Ensure the function you’re integrating is continuous over the interval.
• Pay attention to the limits of integration; they play a key role in the evaluation.
• After using u-substitution, adjust the limits of integration to match the new variable.

In our original exercise, after substituting and rewriting the integral, we use the given information that \(\int_{0}^{x} f(t) dt = 4\) to evaluate the definite integral. By doing so, we establish that the value of the rewritten integral from 0 to 9 is 18.

Continuous Functions

The concept of \textbf{continuous functions} is foundational in calculus, especially for integral calculus. A function is continuous at a point if you can draw its graph at that point without lifting your pen. More formally, a function \(f(x)\) is continuous at a point \(x = a\) if \(\lim_{x \to a} f(x) = f(a)\).
A function must be continuous over the interval of integration for the Fundamental Theorem of Calculus to apply directly. This theorem allows us to compute the definite integral of a function using its antiderivative.

Importance of Continuity in Integration

• It guarantees the existence of an antiderivative required in the evaluation of definite integrals.
• Discontinuities may lead to improper integrals that require special techniques or limits to evaluate.
• The behavior of continuous functions is predictable and easier to analyze.

The problem provided specifies that the function \(f\) is continuous, allowing us to use the Fundamental Theorem of Calculus without concern for points of discontinuity and ensuring the definite integral evaluation is valid for the given interval.