Problem 27 - Applied Calculus

Question

Use a definite integral to find the area under each curve between the given \(x\) -values. For Exercises \(19-24\) also make a sketch of the curve showing the region. $$ f(x)=6 x^{2}+4 x-1 \text { from } x=1 \text { to } x=2 $$

Step-by-Step Solution

Answer

The area under the curve is 19 square units.

Step 1: Understand the Problem

We need to find the area under the curve of the function \( f(x) = 6x^2 + 4x - 1 \) from \( x=1 \) to \( x=2 \). This requires us to use definite integration to compute this area.

Step 2: Set Up the Definite Integral

To find the area under the curve, set up the definite integral of \( f(x) = 6x^2 + 4x - 1 \) from \( x = 1 \) to \( x = 2 \). The integral we need to compute is:\[\int_{1}^{2} (6x^2 + 4x - 1) \, dx\]

Step 3: Integrate the Function

Calculate the indefinite integral of \( 6x^2 + 4x - 1 \). The antiderivative is:\[\int (6x^2 + 4x - 1) \, dx = 2x^3 + 2x^2 - x + C\]where \( C \) is the constant of integration.

Step 4: Evaluate the Definite Integral

Substitute the limits \( x = 1 \) and \( x = 2 \) into the antiderivative to find the definite integral:\[\left[ 2x^3 + 2x^2 - x \right]_{1}^{2} = (2(2)^3 + 2(2)^2 - 2) - (2(1)^3 + 2(1)^2 - 1)\]

Step 5: Calculate Each Part

Compute the evaluated terms for \( x = 2 \):\[2(2)^3 + 2(2)^2 - 2 = 16 + 8 - 2 = 22\]And compute the evaluated terms for \( x = 1 \):\[2(1)^3 + 2(1)^2 - 1 = 2 + 2 - 1 = 3\]

Step 6: Subtract to Find the Area

Subtract the value at \( x = 1 \) from the value at \( x = 2 \) to find the area:\[22 - 3 = 19\]The area under the curve from \( x=1 \) to \( x=2 \) is 19 square units.

Key Concepts

Antiderivative

The concept of an antiderivative, also known as an indefinite integral, is pivotal in understanding calculus. An antiderivative of a function \( f(x) \) is another function \( F(x) \) such that when you take the derivative of \( F(x) \), you get back \( f(x) \). In other words, \( F'(x) = f(x) \). This process is essentially the reverse of differentiation. It tells us the original function \( F(x) \) from which the derivative \( f(x) \) was derived.
Computing the antiderivative involves adding a constant of integration (\( C \)) because the derivative of a constant is zero. Thus, we write the antiderivative as \( F(x) = \int f(x) \, dx + C \). For the given problem, the antiderivative of the function \( 6x^2 + 4x - 1 \) is calculated as \( 2x^3 + 2x^2 - x + C \). The antiderivative provides the basis for evaluating definite integrals and finding the area under a curve.

Area under a curve

Finding the area under a curve between two points on the \( x \)-axis involves calculating the definite integral of the function representing the curve. The definite integral is a fundamental tool in calculus for measuring this area. It is represented as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the bounds or limits of integration.
The definite integral evaluates the accumulation of quantities, in this case, the total area between the curve \( f(x) \) and the \( x \)-axis from \( x = 1 \) to \( x = 2 \). This area can be visualized as a region enclosed by the curve, the \( x \)-axis, and vertical lines at \( x = 1 \) and \( x = 2 \).

• The function \( f(x) = 6x^2 + 4x - 1 \) is integrated within these limits, yielding the area of 19 square units based on the solution.
• A sketch of the curve helps in visualizing the problem and understanding the region being calculated.

The definite integral gives both the algebraic and geometric interpretation of the area under the curve.

Integration by substitution

Integration by substitution is a handy technique for evaluating definite and indefinite integrals, particularly when faced with more complex functions. It is akin to applying the chain rule in reverse. The main idea is to simplify an integral by substituting a part of the integral with a new variable, which helps in making the function easier to integrate.
This technique is often applied when the integrand includes a composition of functions or involves a function and its derivative. The substitution method transforms the original function into a simpler one by using a new variable \( u \) such that \( u = g(x) \). Then, switch the differential \( dx \) to \( du \) using \( du = g'(x) \, dx \).

• Even though the original exercise problem of integrating \( 6x^2 + 4x - 1 \) did not require substitution, recognizing when to use it is key.
• If a function is not directly integrable, substitution could greatly reduce the complexity of the integral.

Once integrated, switching back from \( u \) to \( x \) helps in evaluating the definite integral with respect to the original variable.