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Magazine Chapter 5: Problem 3
Find the integral \(\int(z+1) e^{4 z} d z=\) ___________
Short Answer
Expert verified
\(\int (z+1)e^{4z} dz = \left(\frac{4z+3}{16}\right)e^{4z} + C\)
Step by step solution
01
- Identify the Integral Form
The integral \(\int (z+1) e^{4z} dz\) suggests the use of integration by parts since it is of the form \(\int u dv\).
02
- Choose Functions for Integration by Parts
In integration by parts, we select \(u\) and \(dv\) such that \(\int u \, dv = uv - \int v \, du\). Let \(u = (z+1)\) and \(dv = e^{4z} dz\).
03
- Compute the Differentials
Compute the differential du: \(du = dz\). Compute v by integrating dv: \(dv = e^{4z} dz \implies v = \frac{1}{4} e^{4z}\).
04
- Apply the Integration by Parts Formula
Apply the formula \(\int u \, dv = uv - \int v \, du\): \[ \int (z+1) e^{4z} dz = (z+1) \cdot \frac{1}{4} e^{4z} - \int \left( \frac{1}{4} e^{4z} \right) dz. \]
05
- Integrate the Remaining Integral
Now integrate \(\int \left( \frac{1}{4} e^{4z} \right) dz \): \[ \int \left( \frac{1}{4} e^{4z} \right) dz = \frac{1}{4} \cdot \frac{1}{4} e^{4z} = \frac{1}{16} e^{4z}. \]
06
- Put It All Together
Combine the terms obtained: \[ \int (z+1) e^{4z} dz = \frac{1}{4}(z+1)e^{4z} - \frac{1}{16}e^{4z} + C \] Simplify if possible: \[ \int (z+1) e^{4z} dz = \left( \frac{z+1}{4} - \frac{1}{16} \right) e^{4z} + C = \left( \frac{4z+4-1}{16} \right) e^{4z} + C = \left( \frac{4z+3}{16} \right) e^{4z} + C \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
integral calculus
Integral calculus focuses on the process of finding integrals and antiderivatives. This part of calculus is essential for determining areas under curves, among other applications.
In this specific exercise, we found an antiderivative for the function \((z+1)e^{4z}\).
By integrating, we aim to find the accumulated quantity related to the respective rate function. If you're familiar with differentiation, integration can be seen as the reverse process.
Integration by parts, in this instance, was the chosen method because our integral had the form \(\backslash int u dv\), a product of two functions.
In this specific exercise, we found an antiderivative for the function \((z+1)e^{4z}\).
By integrating, we aim to find the accumulated quantity related to the respective rate function. If you're familiar with differentiation, integration can be seen as the reverse process.
- When you see an integral sign, \(\backslash int \), understand that it's requesting the area under the curve of a particular function or the antiderivative.
- To solve an integral, we often rely on various techniques, like substitution, partial fractions, and integration by parts.
Integration by parts, in this instance, was the chosen method because our integral had the form \(\backslash int u dv\), a product of two functions.
exponential functions
Exponential functions are a type of mathematical function where the variable is in the exponent, typically in the form \(e^{kx}\). These functions grow or decay at constant rates.
Exponential functions are significant in calculus because their derivatives and integrals have unique properties.
In our problem, the term \(e^{4z}\) plays a critical role. Its integral, \(\backslash int e^{4z} dz = \backslashfrac{1}{4} e^{4z}\), is essential in applying the technique of integration by parts. By understanding how to integrate and differentiate exponential functions, we solve more complex integrals.
Exponential functions are significant in calculus because their derivatives and integrals have unique properties.
- For instance, the derivative of \(e^{kx}\) is \(ke^{kx}\), maintaining the exponential form.
- In integrals involving exponential functions, recognizing these properties can simplify the solution.
In our problem, the term \(e^{4z}\) plays a critical role. Its integral, \(\backslash int e^{4z} dz = \backslashfrac{1}{4} e^{4z}\), is essential in applying the technique of integration by parts. By understanding how to integrate and differentiate exponential functions, we solve more complex integrals.
integral techniques
Integral techniques are diverse methods to solve integrals. Among them, Integration by Parts is prominent. This method derives from the product rule of differentiation and offers a useful way to handle products of functions.
The Integration by Parts formula is: \(\backslash int u dv = uv - backslash int v du\).
To effectively use this technique, we:
The choice of \(u\) and \(dv\) can significantly simplify or complicate the integral. With practice, patterns emerge, guiding these choices. In our case, \(u = (z+1)\text\) and \(dv = e^{4z} dz text\). Integrate \(dv \) to get \(v\text\).Then follow the formula to solve the integral without confusion.
The Integration by Parts formula is: \(\backslash int u dv = uv - backslash int v du\).
To effectively use this technique, we:
- Identify parts of the integrand \(u\) and \(dv\).
- Compute \(du\) by differentiating \(u\).
- Compute \(v\) by integrating \(dv\).
- Finally, substitute these into the formula and solve the remaining integral.
The choice of \(u\) and \(dv\) can significantly simplify or complicate the integral. With practice, patterns emerge, guiding these choices. In our case, \(u = (z+1)\text\) and \(dv = e^{4z} dz text\). Integrate \(dv \) to get \(v\text\).Then follow the formula to solve the integral without confusion.
