Chapter 8: Problem 20
Evaluate the integrals in Exercises \(1-24\) using integration by parts. $$ \int t^{2} e^{4 t} d t $$
Short Answer
Expert verified
\( \int t^2 e^{4t} \, dt = \frac{t^2}{4}e^{4t} - \frac{t}{8}e^{4t} + \frac{1}{32}e^{4t} + C \).
Step by step solution
01
Identify Functions for Integration by Parts
In integration by parts, we identify two components from the integral \( \int u \, dv = uv - \int v \, du \). Let's choose \( u = t^2 \) and \( dv = e^{4t} \, dt \). This choice is typically made such that the derivative of \( u \) simplifies the problem.
02
Differentiate and Integrate
Differentiate \( u = t^2 \) to find \( du \). This gives \( du = 2t \, dt \). Integrate \( dv = e^{4t} \, dt \) to find \( v \). This gives \( v = \frac{1}{4}e^{4t} \).
03
Apply Integration by Parts Formula
Apply the integration by parts formula: \( \int u \, dv = uv - \int v \, du \). Substituting the expressions: \( \int t^2 e^{4t} \, dt = t^2 \cdot \frac{1}{4}e^{4t} - \int \frac{1}{4}e^{4t} \, (2t) \, dt \).
04
Simplify and Evaluate the Remaining Integral
Simplify the expression: \( \frac{t^2}{4}e^{4t} - \int \frac{1}{2}t \cdot e^{4t} \, dt \). Again, apply integration by parts to \( \int t e^{4t} \, dt \), with \( u = t \) and \( dv = e^{4t} \, dt \).
05
Differentiate and Integrate Again
Differentiate \( u = t \) to find \( du = dt \). Integrate \( dv = e^{4t} \, dt \) again to find \( v = \frac{1}{4}e^{4t} \).
06
Apply Integration by Parts Formula Again
Substitute into the integration by parts formula: \( \int te^{4t} \, dt = t \cdot \frac{1}{4}e^{4t} - \int \frac{1}{4}e^{4t} \, dt \). That integral simplifies to \( \frac{1}{4}t e^{4t} - \frac{1}{16}e^{4t} \).
07
Combine Results
Combine all parts back together: \( \frac{t^2}{4}e^{4t} - \left( \frac{1}{2} \left( \frac{1}{4}t e^{4t} - \frac{1}{16}e^{4t} \right) \). Further simplify to get: \( \frac{t^2}{4}e^{4t} - \frac{t}{8}e^{4t} + \frac{1}{32}e^{4t} + C \), where \( C \) is the integration constant.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Definite and Indefinite Integrals
In calculus, integrals are a fundamental concept used to find areas under curves, among other things. There are two main types of integrals: definite and indefinite. A definite integral, \( \int_{a}^{b} f(x) \, dx \), gives the net area between the curve and the x-axis, from point \( a \) to point \( b \). It has limits of integration and results in a specific numerical value.
Indefinite integrals, on the other hand, are used to calculate a family of functions. The integral \( \int f(x) \, dx \) finds an entire antiderivative of \( f(x) \); hence, it lacks limits and includes an arbitrary constant \( C \,\).
When solving problems involving indefinite integrals, as in our example, this constant \( C \) emerges because differentiation eliminates any constant.
Indefinite integrals, on the other hand, are used to calculate a family of functions. The integral \( \int f(x) \, dx \) finds an entire antiderivative of \( f(x) \); hence, it lacks limits and includes an arbitrary constant \( C \,\).
When solving problems involving indefinite integrals, as in our example, this constant \( C \) emerges because differentiation eliminates any constant.
- This constant \( C \) is crucial as it accounts for any vertical shifts in the graph of the antiderivative.
- For indefinite integrals, you'll often see results in the form of \( F(x) + C \), where \( F(x) \) is one antiderivative of \( f(x) \).
Exponential Functions
Exponential functions are mathematical functions of the form \( f(t) = a e^{kt} \,\), where \( a \) is a constant, \( e \) is the base of the natural logarithm, and \( k \) is a constant that determines the growth rate. These functions are powerful in modeling real-world scenarios like population growth and radioactive decay.
In our integral example, the function \( e^{4t} \) indicates exponential growth with a rate proportional to \( 4t \).
In our integral example, the function \( e^{4t} \) indicates exponential growth with a rate proportional to \( 4t \).
- The derivative of an exponential function with respect to its variable exponent maintains its exponential form, making it especially useful in calculus for simplifying complex problems.
- An exponential's fundamental property is the rule \( \frac{d}{dt} e^{kt} = ke^{kt} \,\), which becomes handy in integration by parts, as seen in our solution.
Polynomial Functions
Polynomial functions are expressions comprising variables raised to non-negative integer powers, multiplied by coefficients. A typical polynomial function looks like \( f(t) = at^n + bt^{n-1} + \ldots + c \,\).
In solving our integral \( \int t^2 e^{4t} \, dt \,\), the polynomial part is \( t^2 \). Choosing parts in polynomial expressions is crucial for effective integration by parts.
In solving our integral \( \int t^2 e^{4t} \, dt \,\), the polynomial part is \( t^2 \). Choosing parts in polynomial expressions is crucial for effective integration by parts.
- The derivative of a polynomial reduces its power: differentiating \( t^2 \) yields \( 2t \), simplifying steps in our problem.
- When working with polynomials in integrals, look for opportunities to decrease the power through differentiation to make calculus manipulations easier.