Evaluate. (Be sure to check by differentiating!)
[tex]$\int x^{13} e^{x^{14}} d x$[/tex]
A. [tex]$u=x^{13} e^x$[/tex]
B. [tex]$u=x^{14}$[/tex]
C. [tex]$u=x^{13}$[/tex]
D. [tex]$u =e^{ x }$[/tex]
Write the integral in terms of [tex]$u$[/tex].
[tex]$\int x^{13} e^{x^{14}} d x=\int$[/tex] du
(Type an exact answer. Use parentheses to clearly denote the argument of each function.)
Answer (2)
Choose the substitution u = x 14 .
Find the differential d u = 14 x 13 d x .
Express x 13 d x in terms of d u : x 13 d x = 14 1 d u .
Substitute into the integral: ∫ x 13 e x 14 d x = ∫ 14 1 e u d u . The answer is 14 1 e u
Explanation
Problem Analysis We are given the integral ∫ x 13 e x 14 d x and asked to rewrite it in terms of u using an appropriate substitution. We also need to choose the correct substitution from the given options.
Choosing the Substitution Let's analyze the given options for substitution:
Option A: u = x 13 e x . This substitution would make the integral more complex, not simpler. Option B: u = x 14 . This looks promising because the derivative of x 14 is 14 x 13 , which is present in the integral (up to a constant). Option C: u = x 13 . This substitution would not simplify the exponential term. Option D: u = e x . This substitution doesn't account for the x 13 term.
Finding the Differential Based on the analysis, the best choice for substitution is u = x 14 . Now, let's find the differential d u :
d x d u = 14 x 13 d u = 14 x 13 d x
Expressing in terms of du Now, we need to express x 13 d x in terms of d u :
x 13 d x = 14 1 d u
Substituting into the Integral Substitute u = x 14 and x 13 d x = 14 1 d u into the integral:
∫ x 13 e x 14 d x = ∫ e u 14 1 d u = 14 1 ∫ e u d u
Rewriting the Integral Therefore, the integral in terms of u is:
∫ x 13 e x 14 d x = ∫ 14 1 e u d u
Examples
In physics, when dealing with exponential decay or growth processes, integrals of this form often arise. For example, calculating the total amount of a substance produced in a chemical reaction where the rate of production increases exponentially with time involves solving integrals similar to the one given. By using substitution, we can simplify these integrals and find the total production over a specific time period, which is crucial for process optimization and control.
The appropriate substitution for the integral ∫ x 13 e x 14 d x is u = x 14 , leading to the expression ∫ 14 1 e u d u . This simplifies to 14 1 e x 14 + C after substituting back for u . Therefore, the answer is ∫ x 13 e x 14 d x = ∫ 14 1 e u d u .
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