Problem 55 - Calculus Single Variable

Question

\(F(x)\) 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. $$ F(x)=\int_{x}^{1}\left(2 t+1 / t^{2}\right) d t $$

Step-by-Step Solution

Answer

The function \(F(x)\) is given by \(F(x) = -x^2 + \frac{1}{x}\).

Step 1: Identify the Integration Limits

First, observe the given definite integral: \( \int_{x}^{1}(2t+\frac{1}{t^2}) \, dt \). The variable \(x\) is in the lower limit, and \(1\) is in the upper limit.

Step 2: Integrate the Function

Now, integrate the function \(2t+\frac{1}{t^2}\) with respect to \(t\). Begin by integrating each part separately. The integral of \(2t\) is \(t^2\), and the integral of \(\frac{1}{t^2}\) is \(-\frac{1}{t}\). So, the antiderivative is \(t^2 - \frac{1}{t}\).

Step 3: Evaluate the Indefinite Integral

Now, evaluate the antiderivative at the given limits. Substitute the upper limit \(1\): \(1^2 - \frac{1}{1} = 1 - 1 = 0\). Substitute the lower limit \(x\): \(x^2 - \frac{1}{x}\).

Step 4: Apply the Fundamental Theorem of Calculus

According to the Fundamental Theorem of Calculus, the value of the definite integral is the difference between the values of the antiderivative evaluated at the upper and lower limits: \(0 - (x^2 - \frac{1}{x})\).

Step 5: Simplify the Expression

Subtract the lower limit evaluation from the upper limit evaluation to find \(F(x)\). The expression becomes: \(F(x) = -(x^2 - \frac{1}{x})\). Simplifying gives \(F(x) = -x^2 + \frac{1}{x}\).

Key Concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the concepts of differentiation and integration, showing how they are inversely related. It essentially tells us how to evaluate definite integrals using antiderivatives.
The theorem has two main parts, but in the context of definite integrals, we focus on the second part. This part states that if you have a continuous function over an interval and you find its antiderivative, then the definite integral of the function over that interval is the difference of the values of the antiderivative evaluated at the upper and lower limits of the integral.
Think of it as a tool that allows you to calculate the area under a curve (from one point to another) by finding an antiderivative.
Simply put: \[ \int_a^b f(t) \, dt = F(b) - F(a) \] Where \( F(t) \) is any antiderivative of \( f(t) \). This allows students to convert a challenging integration task into a straightforward calculation.

Integration Limits

Definite integrals are evaluated using two specific bounds known as the integration limits. In the function \( F(x) = \int_{x}^{1}(2t + \frac{1}{t^2}) \, dt \), the integration limits are crucial because they define the range over which we compute the area under the curve \( f(t) = 2t + \frac{1}{t^2} \).
The variable \( x \) is the lower limit, while \( 1 \) is the upper limit. This means we're interested in the integral from \( x \) up to \( 1 \).
Integration limits have a significant effect as:

• The order of the limits matters. Switching the limits changes the sign of the integral.
• The limits provide specific values at which the antiderivative is evaluated, explicitly defining the definite integral.

By understanding these limits, students can accurately calculate definite integrals and understand the domain of the integration.

Antiderivative

An antiderivative, sometimes called an indefinite integral, is a function whose derivative is the original function you start with. It's fundamental in finding definite integrals because, according to the Fundamental Theorem of Calculus, the integral of a function is the antiderivative evaluated at two limits.
In our exercise, to solve \( \int_{x}^{1}(2t + \frac{1}{t^2}) \, dt \), we first find an antiderivative for the integrand:

• The antiderivative of \( 2t \) is \( t^2 \).
• The antiderivative of \( \frac{1}{t^2} \) is \( -\frac{1}{t} \).

Combining these results, the antiderivative is \( t^2 - \frac{1}{t} \). This reveals how antiderivatives are used to compute the values of definite integrals. With practice, students will find constructing antiderivatives an invaluable skill in calculus.