Problem 43 - Calculus of a Single Variable

Question

Integration by Substitution In Exercises \(43-46,\) 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

Answer

The evaluated integral is \(2\sqrt{e^{t}-3} + C\)

Step 1: Substitution

Replace \(t\) in the integral in terms of \(u\) with the equation given. If \(u=\sqrt{e^{t}-3}\), square both sides to express \(t\) in terms of \(u\), we get \(u^2 = e^{t}-3\). Adding 3 to both sides gives us \(e^{t}=u^2 + 3\).

Step 2: Derive equation

Find the derivative of \(e^{t}\) respect to \(t\). The derivative of \(e^{t}\) is \(e^{t}\) which equals to \(du/dt = u/2\). Solving for dt gives us \(dt = 2du/u\).

Step 3: Substituting \(dt\)

Substitute \(dt\) with the equation found above in the integral. Our integral is now \(\int u * 2du/u\)

Step 4: Simplify the New Integral

Simplify the integral to \(\int 2du\).

Step 5: Evaluate the Integral

Evaluate the integral to get \(2u + C\).

Step 6: Substitute \(u\) back in

Return back to the \(t\) variable by substituting \(u = \sqrt{e^{t}-3}\) back in. This gives us \(2\sqrt{e^{t}-3} + C\)

Step 7: Final Answer

Our final answer is \(2\sqrt{e^{t}-3} + C\), don't forget the constant of integration!

Key Concepts

Indefinite Integrals

Indefinite integrals represent the family of all antiderivatives of a function. When we look at the expression \(\int f(x)\, dx\), what we are really searching for is a function F(x) whose derivative equals f(x). The primary reason we call this an \'indefinite integral\' is because the solution is not a single function, but an infinite set of functions that differ by a constant, which we denote as C, known as the constant of integration. Therefore, the general solution to an indefinite integral takes the form F(x) + C, where F(x) is any antiderivative of f(x). The process of finding this function requires a solid comprehension of derivative calculus, as it often involves reversing differentiation operations.

Exponential Functions

Exponential functions are characterized by the presence of an exponent that is a variable. One of the most common bases for exponential functions is the mathematical constant e, approximately equal to 2.71828. The function \(e^x\) has a unique property where its derivative is also \(e^x\), making it a particularly important function in calculus. This property simplifies both differentiation and integration processes involving exponentials. In our integration by substitution exercise, \(e^t\) plays a crucial role and understanding how exponential functions behave is essential to solve the integral successfully.

U-substitution

The method of u-substitution is a technique used to simplify integrals by substituting part of the integral with a new variable (typically u), making the integral easier to evaluate. This is analogous to the common strategy in algebra where we make a substitution to simplify an expression or an equation. In the calculus context, u-substitution often requires us to take the derivative of our substitution to substitute for dt, as shown in the given exercise. Once we simplify the integral in terms of u and find the antiderivative, we must remember to substitute back in terms of the original variable to complete the problem.

Derivative Calculus

Derivative calculus, often simply called differentiation, is a fundamental concept in calculus that measures how a function changes as its input changes. It provides a precise way to analyze the rate of change or the slope of the curve of a function at any point. In the context of our integration exercise, we use derivative calculus to find an expression for dt, which allows us to perform the u-substitution correctly. Seeing as the derivative of \(e^t\) is itself, we use this knowledge to solve for dt and then proceed with the substitution and integration. Derivative calculus is not just about calculating slopes; it's an invaluable tool for solving a variety of problems in mathematics and science.