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Magazine Chapter 6: Problem 30
First make a substitution and then use integration by parts to evaluate the integral. $$ \int_{1}^{4} e^{\sqrt{x}} d x $$
Short Answer
Expert verified
The integral evaluates to \( e^2 - 1 \).
Step by step solution
01
Apply Substitution
Let \( u = \sqrt{x} \). Thus, \( x = u^2 \). Find the derivative of \( u \) with respect to \( x \): \( \frac{du}{dx} = \frac{1}{2\sqrt{x}} \implies dx = 2u \, du \). Substitute the expressions into the integral: \[ \int_{1}^{4} e^{\sqrt{x}} \, dx = \int_{1}^{2} e^{u} \, 2u \, du = 2 \int_{1}^{2} ue^{u} \, du \] Change the bounds according to the substitution: When \( x = 1 \), \( u = \sqrt{1} = 1 \). When \( x = 4 \), \( u = \sqrt{4} = 2 \).
02
Setup Integration by Parts Formula
Integration by parts is given by the formula: \( \int u \, dv = uv - \int v \, du \). Here, let \( u = u \) and \( dv = e^{u} \, du \). Then, \( du = du \) and \( v = e^{u} \).
03
Apply Integration by Parts
Apply the integration by parts formula: \( \int ue^{u} \, du = ue^{u} - \int e^{u} \, du \). Evaluate \( \int e^{u} \, du \): It equals \( e^{u} \). Therefore, \[ ue^{u} - \int e^{u} \, du = ue^{u} - e^{u} \].
04
Evaluate the Definite Integral
Now evaluate from \( u = 1 \) to \( u = 2 \): \[ \left[ ue^{u} - e^{u} \right]_{1}^{2} = \left[ 2e^{2} - e^{2} \right] - \left[ e^{1} - e^{1} \right] = e^{2} - 1 \].
05
Write the Final Answer
The integral evaluated from \( x = 1 \) to \( x = 4 \) is: \[ e^{2} - 1 \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integration Techniques
Integration techniques are methods used to solve integrals, which are mathematical expressions representing the area under a curve. There are several methods, and choosing the right one can simplify the process significantly.
One of the fundamental techniques is **integration by parts**. This method is inspired by the product rule of differentiation and is used mainly when the integrand is a product of two functions. It is expressed as follows:
\[ \int u \, dv = uv - \int v \, du \]
Here, choosing which part of the integrand to differentiate and which to integrate is crucial. A good rule of thumb is to use the LIATE rule (Logarithmic, Inverse, Algebraic, Trigonometric, Exponential) to determine which function to pick for \(u\).
Another technique is the **substitution method**, which simplifies the integral by changing variables. This method is particularly useful when dealing with composite functions, as it transforms the integral into a simpler form. By substituting part of the integral with a new variable, the complexity reduces drastically, making it easier to solve.
One of the fundamental techniques is **integration by parts**. This method is inspired by the product rule of differentiation and is used mainly when the integrand is a product of two functions. It is expressed as follows:
\[ \int u \, dv = uv - \int v \, du \]
Here, choosing which part of the integrand to differentiate and which to integrate is crucial. A good rule of thumb is to use the LIATE rule (Logarithmic, Inverse, Algebraic, Trigonometric, Exponential) to determine which function to pick for \(u\).
Another technique is the **substitution method**, which simplifies the integral by changing variables. This method is particularly useful when dealing with composite functions, as it transforms the integral into a simpler form. By substituting part of the integral with a new variable, the complexity reduces drastically, making it easier to solve.
Definite Integrals
Definite integrals calculate the net area under a curve within given bounds. They provide a precise value as opposed to indefinite integrals, which represent a family of functions plus a constant.
Notation for a definite integral is \( \int_{a}^{b} f(x) \, dx \), where \(a\) and \(b\) are the lower and upper limits, respectively. The result of a definite integral between these bounds is a specific number.
In the context of our problem, after the substitution and using integration by parts, the expression becomes an easier definite integral to evaluate. Changing the bounds is an essential step when applying the substitution method, ensuring that the limits reflect the new variable.
After successfully integrating, you apply the limits to the antiderivative to find the final result. This step involves substituting the upper and lower bounds into the resulting function and finding the difference between these values.
Notation for a definite integral is \( \int_{a}^{b} f(x) \, dx \), where \(a\) and \(b\) are the lower and upper limits, respectively. The result of a definite integral between these bounds is a specific number.
In the context of our problem, after the substitution and using integration by parts, the expression becomes an easier definite integral to evaluate. Changing the bounds is an essential step when applying the substitution method, ensuring that the limits reflect the new variable.
After successfully integrating, you apply the limits to the antiderivative to find the final result. This step involves substituting the upper and lower bounds into the resulting function and finding the difference between these values.
Substitution Method
The substitution method is a popular integration technique used to simplify the integration process by transforming the integral into a more manageable form.
In the exercise, the substitution
\( u = \sqrt{x} \)
was made, which transforms the original integral into an equivalent one in terms of \(u\). The derivative \( \frac{du}{dx} = \frac{1}{2\sqrt{x}} \) was used to express \(dx\) in terms of \(du\), leading to
\( dx = 2u \, du \).
This adjustment simplifies the integral, making it easier to handle. It also requires carefully altering the bounds of the integral, as the limits at \(x = 1\) and \(x = 4\) translate into \(u = 1\) and \(u = 2\). This process ensures that the new variable accurately reflects the original integral's limits.
This method is invaluable when dealing with functions that are not straightforward to integrate directly, making a potentially complex problem much more approachable.
In the exercise, the substitution
\( u = \sqrt{x} \)
was made, which transforms the original integral into an equivalent one in terms of \(u\). The derivative \( \frac{du}{dx} = \frac{1}{2\sqrt{x}} \) was used to express \(dx\) in terms of \(du\), leading to
\( dx = 2u \, du \).
This adjustment simplifies the integral, making it easier to handle. It also requires carefully altering the bounds of the integral, as the limits at \(x = 1\) and \(x = 4\) translate into \(u = 1\) and \(u = 2\). This process ensures that the new variable accurately reflects the original integral's limits.
This method is invaluable when dealing with functions that are not straightforward to integrate directly, making a potentially complex problem much more approachable.
