Problem 11 - Calculus Single Variable

Question

In Exercises \(11-20,\) calculate \(F(x)=\int_{a}^{x} f(t) d t\) $$ f(t)=3 t^{2}+1 \quad a=-1 $$

Step-by-Step Solution

Answer

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

Step 1: Identify the Integral Expression

We are given the function \( f(t) = 3t^2 + 1 \) and we need to compute \( F(x) = \int_{-1}^{x} f(t) \, dt \). This means we will integrate the function \( f(t) = 3t^2 + 1 \) from \( t = -1 \) to \( t = x \).

Step 2: Find the Antiderivative

To find the antiderivative of \( f(t) = 3t^2 + 1 \), we integrate each term of the function separately. For the term \( 3t^2 \), the antiderivative is \( t^3 \), and for the term \( 1 \), it is \( t \). Therefore, the antiderivative of \( f(t) \) is \( T(t) = t^3 + t + C \), where \( C \) is the constant of integration.

Step 3: Evaluate the Indefinite Integral

Replace \( T(t) = t^3 + t \) in the expression for \( F(x) \), as we only need the definite integral without the constant here. The expression becomes \[ F(x) = T(x) - T(-1) = (x^3 + x) - ((-1)^3 + (-1)) \].

Step 4: Simplify the Expression

Calculate \( T(-1) \). We find that \( (-1)^3 + (-1) = -1 + (-1) = -2 \). Substitute back to find \( F(x) = x^3 + x - (-2) = x^3 + x + 2 \).

Key Concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus links the concept of differentiation with integration, two of the main concepts in calculus. This theorem is incredibly powerful because it shows that every continuous function has an antiderivative and provides a way to evaluate definite integrals. When we talk about definite integrals, we are essentially talking about finding the net area under the curve of a function from one point to another. The theorem has two parts:

• First Part: It assures us that the antiderivative of a function, when evaluated at its limits, gives the area under the curve.
• Second Part: It states that if you have a continuous function, then you can find another function whose derivative is the original function.

In our exercise, we used the Fundamental Theorem of Calculus to compute the definite integral of the function \(f(t) = 3t^2 + 1\) over a specific interval \([-1, x]\). By finding the antiderivative and evaluating it at the bounds of the integral, we arrived at the solution.

Antiderivative

An antiderivative of a function is another function, derived by reversing the process of differentiation. If you know how to differentiate, then finding an antiderivative is essentially working backwards.

• If \(f(t)\) is a function, its antiderivative \(F(t)\) satisfies the condition \(F'(t) = f(t)\).
• To find antiderivatives, you essentially integrate the function, adding a constant \(C\) because differentiation of a constant results in zero, leaving many possible antiderivatives.

In our example, we have the function \(f(t) = 3t^2 + 1\). By finding the antiderivative, we got \(T(t) = t^3 + t + C\). This is important because the antiderivative is used in solving the definite integral.

Polynomial Integration

Polynomial integration is a method used to find the integral of polynomial functions. Polynomials are an important class of functions in calculus and appear often in real-world applications.

• The basic rule for integrating a polynomial \(ax^n\) is to increase the power by one and divide by the new power: \(\int ax^n dx = \frac{a}{n+1}x^{n+1} + C\).
• When integrating a polynomial, each term is integrated separately, making calculations straightforward.

In the exercise at hand, the function \(f(t) = 3t^2 + 1\) was integrated term by term:

• The term \(3t^2\) becomes \(t^3\) after integration.
• The constant term \(1\) becomes \(t\).

So, the antiderivative was \(t^3 + t\), which was crucial in finding the definite integral from \(t = -1\) to \(t = x\). This approach to polynomial integration simplifies the process of finding antiderivatives and evaluating definite integrals.