Question

Definite integrals Evaluate the following integrals using the Fundamental Theorem of Calculus. $$\int_{1}^{2} \frac{3}{t} d t$$

Short answer

Answer: The value of the definite integral is $$3\ln{2}$$.

Step-by-step solution

Determine the anti-derivative of the function

First, we must find the anti-derivative of the function inside the integral. We have $$\frac{3}{t}$$ To find its anti-derivative, we use the basic formula: $$\int \frac{a}{t} dt = a \ln|t| + C$$ In our case, \(a = 3\). Therefore, the anti-derivative is $$3\ln|t| + C$$

Apply the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus tells us that if we have a continuous function \(f\) and want to find the integral over the interval \([a, b]\), we can use the anti-derivative, \(F\). The theorem states: $$\int_{a}^{b} f(t) dt = F(b) - F(a)$$ In our case, we have the antiderivative \(F(t)=3\ln|t|+C\), and want to evaluate the integral for the interval \([1, 2]\). Therefore, we can substitute the values of \(a=1\) and \(b=2\) into the formula: $$\int_{1}^{2} \frac{3}{t} dt = (3\ln|2|+C) - (3\ln|1|+C)$$

Evaluate the expression

Now we have: $$\int_{1}^{2} \frac{3}{t} d t = (3\ln{2}+C) - (3\ln{1}+C)$$ Since \(\ln{1} = 0\), the expression simplifies to: $$\int_{1}^{2} \frac{3}{t} d t = (3\ln{2}+C) - (0+C)$$ Notice that the constant \(C\) will cancel itself out: $$\int_{1}^{2} \frac{3}{t} d t = 3\ln{2}$$ So, the value of the integral is: $$\int_{1}^{2} \frac{3}{t} d t = 3\ln{2}$$

Key concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the world of derivatives and integrals, which are two primary operations in calculus.
This theorem provides a beautiful link between the concept of a function's derivative and its antiderivative.
There are two parts to this theorem:

• **First Part**: This part of the theorem states that if you have a continuous function and you take its integral, then the derivative of this integral is the original function.
  Essentially, differentiation and integration are inverse processes.
• **Second Part**: This is the one we often use to evaluate definite integrals, as we did in the example.
  It tells us that to find the definite integral of a function over an interval, you can take the antiderivative at the endpoint of the interval and subtract the antiderivative at the beginning of the interval.
  Mathematically, \[\int_{a}^{b} f(t) \, dt = F(b) - F(a)\]

In practice, this means once we find the antiderivative (as we did with \(F(t) = 3 \ln |t|\)), we just need to substitute the bounds to find the definite integral.
This theorem simplifies the process of solving definite integrals significantly.

Antiderivative

An antiderivative, sometimes known as an indefinite integral, is a function that reverses the process of differentiation.
If you differentiate the antiderivative, you will get back the original function.
In essence, finding an antiderivative involves finding a function whose derivative is the original function being integrated.
For example, in our problem, we started with the function \(\frac{3}{t}\).
The antiderivative of this function is \(3 \ln |t|\) (ignoring the constant of integration \(C\), as it cancels out in definite integrals).
Typically, finding antiderivatives involves using basic integration rules, such as:

• Integrating \(x^n\) results in \(\frac{x^{n+1}}{n+1}+C\), where \(n eq -1\).
• The antiderivative of \(1/x\) is \(\ln |x| + C\), as used in the example.
• For exponentially or trigonometric functions, use specific formulas like integration of \(e^x\) results in \(e^x + C\).

Mastering the finding of antiderivatives is key to solving problems involving definite integrals efficiently.

Continuous Function

A continuous function is one where there are no breaks, jumps, or discontinuities within its domain.
In simple terms, you can draw the graph of a continuous function without lifting your pencil from the paper.
Continuity is an important requirement for the application of the Fundamental Theorem of Calculus.
This is because both parts of the theorem assume the function being integrated is continuous over the interval in question.
Characteristics of continuous functions include:

• No gaps: The function value is defined for every point in the interval.
• No jumps: As you approach any point \(c\) in the interval from either direction, the values get arbitrarily close to the function value at \(c\).
• No vertical asymptotes: The function does not approach infinity within the interval.

In our example, \(\frac{3}{t}\) is continuous over the interval \(1 \leq t \leq 2\) since there are no breaks or undefined points in the specified range.
This ensures that using the Fundamental Theorem is valid.