Question

Evaluate the following integrals. $$\int_{0}^{2} 4 x e^{-x^{2} / 2} d x$$

Short answer

Question: Evaluate the integral $\int_{0}^{2} 4xe^{-x^2/2}dx$. Answer: The value of the integral is $-4(e^2 - 1)$.

Step-by-step solution

Identify the substitution

We can identify that the integral has a composite function, where the function $$e^{-x^2/2}$$ has an inner function of $$-x^2/2$$. We will use substitution with $$u = -\frac{x^2}{2}$$, so that $$du = -xdx$$.

Rewrite the integral in terms of u

To rewrite the integral in terms of $$u$$, we first need to solve for $$dx$$, which is done by dividing both sides of our expression, $$du = -xdx$$, by $$-x$$. We obtain $$dx = -\frac{du}{x}$$. Also, change the limits of integration from $$x$$ to $$u$$: When $$x=0$$, $$u = -\frac{0^2}{2} = 0$$ When $$x=2$$, $$u = -\frac{2^2}{2} = -2$$ The new integral, in terms of $$u$$, is: $$\int_{0}^{-2} 4xe^{-u}\left(-\frac{du}{x}\right)$$

Simplify the integral and solve

The integral can now be simplified as: $$- 4\int_{0}^{-2} e^{-u}\, du$$ Now, we can find the antiderivative of $$e^{-u}$$ which is -$$e^{-u}$$. Therefore, the integral becomes: $$- 4 [-e^{-u}]_{0}^{-2}$$

Evaluate the definite integral using the Fundamental Theorem of Calculus

By applying the Fundamental Theorem of Calculus, we have: $$ -4[-e^{-(-2)} - -e^{-(0)}] = -4[e^2 - 1]$$

Simplify the final expression

The final result for the given definite integral is: $$-4(e^2 - 1)$$

Key concepts

Substitution Method

The substitution method is a useful technique for solving integrals, especially when you have a composite function, meaning a function within another function. Imagine peeling an onion, where each peel looks like another layer of the onion. The substitution method allows us to make the integral process easier by substituting a part of the integrand that is complex with a single variable, typically denoted as "u".

• We identify the inner function, here it's \( -\frac{x^2}{2} \), and then set \( u = -\frac{x^2}{2} \).
• Then, differentiate \( u \) to find \( du \): \(du = -x dx\).
• This step transforms the integral's variable into terms of \( u \), making it simpler to evaluate.

This is akin to changing enclosures to make a math problem more manageable. Using this method, you can efficiently solve otherwise complicated integrals by essentially converting the problem into a simpler one. By simplifying the function, we make integration more straightforward, as seen in the exercise.

Definite Integrals

A definite integral is an integral that calculates the net area under a curve over an interval \( [a, b] \). The notation \( \int_{a}^{b} f(x) \, dx \) indicates this calculation.

For definite integrals, the limits of integration \( a \) and \( b \) are applied after finding the indefinite integral, which gives an exact numerical answer rather than a general formula with an at least one constant.

• Definite integrals help in calculating areas, volumes, and more in practical applications.
• When performing transformations, like with substitution, you must also transform these limits accordingly. For example, switch from \( x \) to \( u \) values to remain consistent.

This idea is pivotal in definite integrals: they provide a concrete result rather than a general expression. They let us measure something specific within set parameters, as demonstrated in the exercise provided.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the concepts of differentiation and integration, showcasing that they are essentially reverse processes.

• The first part of the theorem states that if a function is continuous over \( [a, b] \), then we can find its indefinite integral, which is a new function that, if differentiated, returns the original function.
• The second part is more directly applicable here: it allows us to compute a definite integral once an antiderivative of the function is known. That is, \( \int_{a}^{b} f(x) dx = F(b) - F(a) \), where \( F(x) \) is the antiderivative of \( f(x) \).

In the exercise, we simplified the integral to a form where its antiderivative was easily recognizable, allowing us to use this theorem effectively. After simplifying the integral and recognizing \( -e^{-u} \) as the antiderivative, we employed the theorem to evaluate it over a specific interval, finding the net area under the curve, which is pivotal in calculus problems.