Question

Evaluate the definite integral. $$ \int_{0}^{4} \frac{x}{e^{x / 2}} d x $$

Short answer

The definite integral of the function over the interval [0, 4] is \(- 4 e^{-2} - 4\).

Step-by-step solution

Identify u and dv

According to the LIATE rule, \(u = x\) (Algebraic) and \(dv = \frac{1}{e^{x/2}} dx\) (Exponential). Then calculate \(du = dx\) and integrate \(dv\) to find \(v = -2e^{-x/2}\).

Apply the Integration by Parts formula

With \(u\), \(du\), \(v\), and \(dv\) identified, the integration by parts formula can be applied. The standard integration by parts formula is \(∫udv = uv - ∫vdu\). Thus, the equation becomes \(x*(-2e^{-x/2}) - ∫(-2e^{-x/2})dx \). Then distribute the \(x\) to simplify it into: \(-2x e^{-x/2} - ∫(-2e^{-x/2})dx\).

Solve remaining integral

The remaining integral, \(-2 ∫ e^{-x/2} dx\), can be solved by a simple substitution, like \(z = -x/2\). After solving, this part of the integral becomes \(4e^{-x/2}\).

Combine the solutions

The solution to the integral is the combination of the solutions from Step 2 and Step 3. That is: \(-2x e^{-x/2} - 4 e^{-x/2}\)

Evaluate the definite Integral

To evaluate the integral from 0 to 4, you would replace \(x\) by 4 and then by 0 and subtract the two results. The final answer becomes \(- 4 e^{-2} - 4\)

Key concepts

Integration by Parts

Integration by parts is a technique used to evaluate integrals where the integrand is a product of two functions that cannot be easily integrated as is. It is based on the product rule for differentiation. The general formula for integration by parts is \[ \int u dv = uv - \int v du \.\] To apply this method, one must identify parts of the integrand as 'u' and 'dv'. Then, differentiate 'u' to get 'du' and integrate 'dv' to obtain 'v'. The original integral is then transformed into the product of 'u' and 'v' minus the integral of 'v du'. This strategy can turn a seemingly complicated integral into a simpler one, or into a form where other integration techniques can be applied.

When selecting 'u' and 'dv', it's typically helpful to choose 'u' to be a function that becomes simpler when differentiated, and 'dv' to be a function that does not become more complicated when integrated. It’s a powerful method and is particularly useful when dealing with the product of polynomial and exponential functions, logarithms, or trigonometric functions.

LIATE Rule

The LIATE rule is a mnemonic device that helps in selecting 'u' and 'dv' when using integration by parts. LIATE stands for Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, and Exponential functions, in the order of precedence for choosing 'u'. For instance, if an integrand contains an algebraic expression (A) and an exponential function (E), according to LIATE, 'u' should be the algebraic part because A comes before E in the LIATE sequence.

This rule is not a guaranteed method but serves as a useful guideline. Depending on the specifics of the integral, you might find exceptions or need to apply the integration by parts multiple times. Still, for many scenarios, LIATE provides a quick way to make a choice that will simplify the integral.

Exponential Functions

Exponential functions have the form \( f(x) = a^x \), where 'a' is a positive constant. In calculus, the most frequently encountered exponential function is the natural exponential function, where \( a = e \), the base of natural logarithms, approximately equal to 2.71828. This function is denoted as \( e^x \).

Integrating exponential functions often requires the use of substitution or integration by parts, especially when combined with other types of functions, such as polynomial or trigonometric functions. Exponential functions have the interesting property that their derivative and integral are proportional to themselves, which greatly influences their behavior when analyzed in the context of calculus.

U-Substitution

U-substitution is a method in calculus for finding the integral of a composed function whose derivative is also present in the integrand. This technique is an inverse chain rule, aiming to simplify the integration process. The basic idea behind u-substitution is to choose a new variable 'u', which is a function of 'x', such that when substituted into the integral, it cancels out parts of the original integrand, making it easier to integrate.

After choosing an appropriate 'u', one must compute 'du', which is the derivative of 'u' with respect to 'x', and proceed to express the original integral in terms of 'u' and 'du'. After integrating with respect to 'u', it is usually necessary to substitute back in order to express the integral in terms of the original variable 'x'. U-substitution works particularly well when there is a function and its derivative multiplied together within an integral.