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Magazine Chapter 9: Problem 40
Evaluate $$\int x^{7} e^{x^{4}} d x$$.
Short Answer
Expert verified
\textstyle \frac{1}{4} e^{x^4} (x^4 - 1) + C \.
Step by step solution
01
Identify the integral to be solved
The integral to be evaluated is \(\textstyle \int x^{7} e^{x^{4}} d x \), which involves the product of a polynomial and an exponential function.
02
Look for a substitution
To simplify the integral, use a substitution that can transform it into an easier form. Let \(\textstyle u = x^4 \). Then, differentiate both sides with respect to \(x\): \(\textstyle \frac{d u}{d x} = 4x^3 \). Thus, \(\textstyle d u = 4x^3 d x \).
03
Adjust the integral for the substitution
Rewrite \(\textstyle dx \) in terms of \(\textstyle du \): \(\textstyle d x = \frac{d u}{4x^3} \). Next, note that \(x^7 = x^4 \cdot x^3 = u x^3 \). Substituting these values into the original integral, we get: \(\textstyle \int x^7 e^{x^4} d x = \int u x^3 e^u \frac{d u}{4x^3} \).
04
Simplify the integral
Notice that \(x^3 \) cancels out, simplifying the integral to: \(\textstyle \int \frac{u e^u}{4} d u = \frac{1}{4} \int u e^u d u \).
05
Use integration by parts
Recall the formula for integration by parts: \(\textstyle \int u v' d x = u v - \int u' v d x \). Let \(v' = e^u \) and \(u' = du \). Then, \(u = u \) and \(v = e^u \). Applying the formula, we get: \(\textstyle \frac{1}{4} \left[ u e^u - \int e^u d u \right] \).
06
Integrate and simplify
Since \( \int e^u d u = e^u \), substitute back to get: \(\textstyle \frac{1}{4} \left[ u e^u - e^u \right] = \frac{1}{4} e^u (u - 1) \). Finally, substitute back \(u = x^4 \) to obtain: \(\textstyle \frac{1}{4} e^{x^4} (x^4 - 1) + C \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integration by Parts
Integration by parts is a method used to integrate the product of two functions. The formula for integration by parts is given by \(\int u \frac{dv}{dx}dx = uv - \int v \frac{du}{dx}dx\). This method is especially useful when the integral involves the product of polynomial and exponential functions or logarithmic functions. In general, you will:
- Choose which part of the integral to set as \(u\) and which as \(dv\).
- Differentiate \(u\) to find \(du\).
- Integrate \(dv\) to find \(v\).
- Substitute and simplify using the integration by parts formula.
Substitution Method
The substitution method, or \(u\)-substitution, is a technique to simplify integrals by changing variables. It's similar to reversing the chain rule for differentiation. The general steps are:
- Identify a substitution \(u = g(x)\) that simplifies the integral. Write \(du = g'(x)dx\).
- Rewrite the original integral in terms of \(u\) and \(du\).
- Integrate with respect to \(u\).
- Substitute back the original variable integration.
Exponential Function
Exponential functions are of the form \(f(x) = e^x\), where \(e\) is a constant approximately equal to 2.71828. They have unique properties such as:
- The derivative of \(e^x\) is \(e^x\).
- The integral of \(e^x\) is also \(e^x\).
