Problem 53 - Calculus Single Variable

Question

In each of Exercises \(53-58, F(x)\) is a function of a variable \(x\) that appears in a limit (or in the limits) of integration of a given definite integral. Express \(F(x)\) explicitly by calculating the integral. $$ F(x)=\int_{1}^{x}\left(3 t^{2}+1\right) d t $$

Step-by-Step Solution

Answer

\( F(x) = x^3 + x - 2 \)

Step 1: Identify the Integral

The function \( F(x) \) is given as a definite integral with the variable \( x \) as the upper limit: \[ F(x) = \int_{1}^{x} (3t^2 + 1) \, dt \]. Here, \( 3t^2 + 1 \) is the integrand, and we aim to find \( F(x) \) by evaluating this integral.

Step 2: Integrate the Function inside the Integral

To find \( F(x) \), we need to calculate the antiderivative of the integrand \( 3t^2 + 1 \). The antiderivative of \( 3t^2 \) is \( t^3 \), and the antiderivative of \( 1 \) is \( t \). So, the antiderivative of \( 3t^2 + 1 \) is \[ t^3 + t + C \], where \( C \) is the constant of integration.

Step 3: Evaluate the Definite Integral

Substitute the limits of integration into the antiderivative to evaluate the definite integral. Calculate: \[ F(x) = \left[ t^3 + t \right]_1^x = \left( x^3 + x \right) - \left( 1^3 + 1 \right) \].

Step 4: Simplify the Expression

Perform the subtraction from the previous step: \[ F(x) = (x^3 + x) - (1 + 1) = x^3 + x - 2 \]. Hence, \( F(x) \) is explicitly expressed as the function \( F(x) = x^3 + x - 2 \).

Key Concepts

Understanding Antiderivatives

An antiderivative is at the heart of integration. When we talk about the antiderivative, we refer to a function whose derivative is the given function. In simpler terms, if you have a function that describes the rate of change, the antiderivative provides a function for the quantity that is changing.
To find the antiderivative of a polynomial, we apply the reverse of differentiation.

• For instance, the antiderivative of a power function like \( t^n \) is \( \frac{t^{n+1}}{n+1} \), provided \( n eq -1 \).
• For the integrand \( 3t^2 + 1 \), we calculate the antiderivative of each part separately.
• The antiderivative of \( 3t^2 \) is \( t^3 \), and the antiderivative of \( 1 \) is \( t \).

Together, the antiderivative of \( 3t^2 + 1 \) becomes \( t^3 + t + C \), where \( C \) is a constant. This constant becomes irrelevant in definite integration, as it cancels out during the evaluation at the limits.

Exploring Limits of Integration

In definite integration, the limits of integration signify the interval over which you evaluate the antiderivative. These are the numbers at the bottom and top of the integral sign, often called the lower and upper limits of integration. Here, the limits are 1 and \( x \).
When you evaluate a definite integral, you replace the variable with the upper limit, then subtract the result of replacing the variable with the lower limit:

• For \( F(x) = \int_{1}^{x} (3t^2 + 1) \, dt \), we plug \( t=x \) into the antiderivative \( t^3 + t \) to get \( x^3 + x \).
• Next, substitute \( t = 1 \) into the antiderivative to calculate the lower end, yielding \( 1^3 + 1 \), which simplifies to 2.
• Finally, subtract the result of the lower limit from the upper: \( (x^3 + x) - (1 + 1) \).

This subtraction gives us \( x^3 + x - 2 \), a clear expression that characterizes \( F(x) \) over \([1, x]\).

The Role of Calculus in Integration

Calculus, with its branches of differential and integral calculus, helps us understand how quantities change over time, space, and other dimensions. Integration, a key concept of calculus, is essentially about accumulating, summing, or combining small quantities.
Within calculus, a definite integral computes the net "area" between a function and the \( x \)-axis over an interval. It can be interpreted as the total accumulation of quantities described by a given rate.

• In our example, the function \( 3t^2 + 1 \) represents a rate of change, and when we integrate it definitively from 1 to \( x \), we find a net accumulation function, \( x^3 + x - 2 \).
• Calculus uses the fundamental theorem of calculus to relate differentiation and integration.
• This theorem states that if \( F \) is an antiderivative of \( f \) over an interval, then the definite integral of \( f \) from \( a \) to \( b \) is \( F(b) - F(a) \).

This powerful relationship allows calculus to solve complex problems in physics, engineering, and beyond, just as it transforms the integral \( \int_{1}^{x} (3t^2 + 1) \, dt \) into a simple function \( x^3 + x - 2 \).