Problem 97 - Thomas Calculus

Question

Evaluate the integrals. \begin{equation}\int 3 x^{\sqrt{3}} d x\end{equation}

Step-by-Step Solution

Answer

\( \int 3x^{\sqrt{3}} \, dx = \frac{3x^{\sqrt{3} + 1}}{\sqrt{3} + 1} + C \)

Step 1: Identify the Integral Formula

The integral we need to evaluate is \( \int 3x^{\sqrt{3}} dx \). We can use the power rule for integration, which is \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), where \( n eq -1 \).

Step 2: Apply the Power Rule

The integrand \( 3x^{\sqrt{3}} \) can be integrated by applying the power rule. Here, \( n = \sqrt{3} \) and the constant multiplier is \( 3 \). First, integrate \( x^{\sqrt{3}} \):\[\int x^{\sqrt{3}} \, dx = \frac{x^{\sqrt{3} + 1}}{\sqrt{3} + 1} + C\]Multiply the result by 3:

Step 3: Multiply by the Constant

After integrating using the power rule, multiply by the constant factor which is 3:\[3 \cdot \left(\frac{x^{\sqrt{3} + 1}}{\sqrt{3} + 1}\right) = \frac{3x^{\sqrt{3} + 1}}{\sqrt{3} + 1} + C\]This is the integrated result of the function.

Key Concepts

Definite Integral

A definite integral is a fundamental concept in calculus that computes the net area under a curve within a specific interval on the x-axis. It provides a precise numerical value representing the accumulation of quantities, such as distance or area, between the two boundaries. Unlike an indefinite integral, which represents a family of functions, a definite integral yields a single number. It is denoted by the integral sign with the lower and upper limits of integration, for example, \[ \int_{a}^{b} f(x) \, dx \]where \( a \) and \( b \) are the limits of integration.The process to compute definite integrals involves finding the antiderivative of the function \( f(x) \), and then evaluating it at the upper limit \( b \) and lower limit \( a \), subtracting the lower value from the upper value. This calculation is important when you're interested in finding specific values such as the exact area between a curve and the x-axis between two points 'a' and 'b'. Understanding definite integrals can immensely help in solving real-world problems involving areas, volumes, and average values.

Power Rule

The power rule is a handy technique for integrating functions where the variable has an exponent. It simplifies the process of integration when the function is in the form of \( x^n \). The power rule formula is: \[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \]for \( n eq -1 \). This formula comes in very handy, particularly when dealing with polynomials.To apply the power rule in practice:

• Identify the exponent \( n \) in the function \( x^n \).
• Increase the exponent by 1 to get \( n + 1 \).
• Divide the new expression \( x^{n+1} \) by the new exponent \( n+1 \).
• Add the constant of integration, \( C \), to represent an entire family of antiderivatives.

This method works swiftly to solve many integration problems, allowing you to find the integral of polynomial expressions efficiently. The power rule is emphasized heavily in calculus due to its wide applicability in both theoretical and practical scenarios.

Indefinite Integral

An indefinite integral represents a family of functions and is the reverse process of differentiation. When we integrate a function without specific bounds, we describe it with an indefinite integral. This is useful to find the general form of antiderivatives for a given function. The general notation for an indefinite integral is:\[ \int f(x) \, dx = F(x) + C \]where \( F(x) \) is the antiderivative of \( f(x) \) and \( C \) is an arbitrary constant.The reason for the constant \( C \) is crucial because when you differentiate the antiderivative \( F(x) \), you lose constant information; thus, \( C \) accounts for all possible vertical shifts in the graph of the function.To compute an indefinite integral:

• Select the appropriate integration technique based on the form of \( f(x) \).
• Apply the technique to find the antiderivative \( F(x) \).
• Add the constant of integration \( C \) to complete the expression.

Understanding how to calculate indefinite integrals is crucial, as it paves the way for solving more complex problems and aids in understanding the fundamental theorem of calculus, connecting integration and differentiation.