Problem 31
Question
Use the specified substitution to find or evaluate the integral. $$ \begin{array}{l} \int \sqrt{e^{t}-3} d t \\ u=\sqrt{e^{t}-3} \end{array} $$
Step-by-Step Solution
Verified Answer
The integral of \(\sqrt{e^{t}-3}\) with respect to \(t\), given the substitution \(u=\sqrt{e^{t}-3}\), is \(1/2 * ln|\sqrt{e^{t}-3}| + C\).
1Step 1: Find the substitution derivative
We have that \(u=\sqrt{e^{t}-3}\). Squaring both sides yields \(u^2=e^{t}-3\). When we add 3 to both sides and take the logarithm (base \(e\)) of both sides, we get \(e^{t}=u^2+3\). This lets us find the derivative \(dt\) in terms of \(du\). The derivative of \(e^{t}\) with respect to \(t\) is \(e^{t}\), which would yield \(dt = du/(2u)\).
2Step 2: Express the integral in terms of u
Substitute \(u\) and \(dt\) into the integral. The integral transforms from \(\int \sqrt{e^{t}-3} dt\) to \(\int u * du/(2u)\).
3Step 3: Simplify and Compute Integral
The \(u\) terms cancel out, yielding \(\int 1/(2u) du\). To find the integral, integrate with respect to \(u\) which yields \(1/2 * ln|u| + C\).
4Step 4: Substitute back for t
Substitute back for \(t\) by replacing \(u\) by \(\sqrt{e^{t}-3}\) in the expression obtained. This gives the final result as \(1/2 * ln|\sqrt{e^{t}-3}| + C\).
Key Concepts
Integral CalculusSubstitution MethodNatural Logarithm
Integral Calculus
Integral calculus is a branch of calculus focused on finding the integral of a function. Essentially, integration is the reverse process of differentiation. It helps us determine the accumulation of quantities, such as areas under curves. If you think about the slope of a function's curve, integration allows us to "accumulate" those slopes over an interval to find areas or other cumulative measures.
In the given exercise, we need to find the integral of the function \(\sqrt{e^t-3}\). Using integration by substitution, a key concept in integral calculus, simplifies the process greatly. Instead of dealing with the complexities directly, we use substitution to transform the integral into a simpler form, facilitating easier computation. This technique highlights the core idea behind integral calculus: transforming a difficult problem into one that is more manageable.
In the given exercise, we need to find the integral of the function \(\sqrt{e^t-3}\). Using integration by substitution, a key concept in integral calculus, simplifies the process greatly. Instead of dealing with the complexities directly, we use substitution to transform the integral into a simpler form, facilitating easier computation. This technique highlights the core idea behind integral calculus: transforming a difficult problem into one that is more manageable.
Substitution Method
The substitution method is a powerful technique in integral calculus used to simplify finding integrals. It involves substituting elements of the function with a new variable, making the integration easier to handle. Our main goal is to transform the integral into a simpler form by introducing a new variable called \(u\).
In our example, we set \(u = \sqrt{e^t-3}\). This substitution simplifies the original function, turning it into something easier to integrate. By squaring both sides, we derived \(u^2 = e^t - 3\), and subsequently, \(e^t = u^2 + 3\) to express \(t\) in terms of \(u\). One of the critical steps is finding the differential \(dt\) in terms of \(du\), resulting in \(dt = du/(2u)\).
In our example, we set \(u = \sqrt{e^t-3}\). This substitution simplifies the original function, turning it into something easier to integrate. By squaring both sides, we derived \(u^2 = e^t - 3\), and subsequently, \(e^t = u^2 + 3\) to express \(t\) in terms of \(u\). One of the critical steps is finding the differential \(dt\) in terms of \(du\), resulting in \(dt = du/(2u)\).
- Transform the integral using substitution: \(\int \sqrt{e^t-3} dt \rightarrow \int u \cdot \frac{du}{2u}\)
- Simplify to \(\int \frac{1}{2} du\)
- Compute the integral and then reverse the substitution
Natural Logarithm
The natural logarithm, denoted as \(ln\), is a logarithm to the base \(e\), where \(e\) is an irrational constant approximately equal to 2.71828. In calculus, the natural logarithm is particularly important because of its straightforward derivative and integral properties.
In our solution, the natural logarithm appears when simplifying the integral \(\int \frac{1}{2u} du\). The integral of \(\frac{1}{u}\) is \(ln|u|\), a fundamental result in calculus. Thus, upon integrating, we obtain \(\frac{1}{2} ln|u| + C\), where \(C\) is an integration constant.
The final step in our solution involves reversing the substitution, giving us \(\frac{1}{2} ln|\sqrt{e^t-3}| + C\). This highlights the seamless transition from substitution to using the natural logarithm as a means to achieve the final integral's expression. Understanding and applying the natural logarithm's properties can greatly simplify many integration problems.
In our solution, the natural logarithm appears when simplifying the integral \(\int \frac{1}{2u} du\). The integral of \(\frac{1}{u}\) is \(ln|u|\), a fundamental result in calculus. Thus, upon integrating, we obtain \(\frac{1}{2} ln|u| + C\), where \(C\) is an integration constant.
The final step in our solution involves reversing the substitution, giving us \(\frac{1}{2} ln|\sqrt{e^t-3}| + C\). This highlights the seamless transition from substitution to using the natural logarithm as a means to achieve the final integral's expression. Understanding and applying the natural logarithm's properties can greatly simplify many integration problems.
Other exercises in this chapter
Problem 30
Sketch the graphs of the function \(g(x)=f(x)+C\) for \(C=-2, C=0,\) and \(C=3\) on the same set of coordinate axes. $$ f(x)=\sqrt{x} $$
View solution Problem 31
In Exercises 31 and \(32,\) show that the function satisfies the differential equation. \(y=a \sinh x\) \(y^{\prime \prime \prime}-y^{\prime}=0\)
View solution Problem 31
In Exercises 31-36, evaluate the integral using the following values. $$\int_{2}^{4} x^{3} d x=60, \quad \int_{2}^{4} x d x=6, \quad \int_{2}^{4} d x=2$$ $$ \in
View solution Problem 31
In Exercises \(31-36,\) find the area of the region bounded by the graphs of the equations. $$ y=3 x^{2}+1, \quad x=0, \quad x=2, \quad y=0 $$
View solution