Problem 19 - Calculus Single Variable

Question

In Exercises \(11-20,\) calculate \(F(x)=\int_{a}^{x} f(t) d t\) $$ f(t)=15 \sqrt{t}(1+t) \quad a=1 \quad(0

Step-by-Step Solution

Answer

The integral \( F(x) = 10x^{3/2} + 6x^{5/2} - 16 \).

Step 1: Identify the Given Function

We are given the function to integrate, which is \( f(t) = 15 \sqrt{t}(1+t) \). We also know that \( a = 1 \), and we need to find \( F(x) = \int_{1}^{x} 15 \sqrt{t}(1+t) \, dt \).

Step 2: Break Down the Function

The expression \( 15 \sqrt{t}(1 + t) \) can be expanded to \( 15 (\sqrt{t} + t^{3/2}) \). So, the function becomes \( 15 \sqrt{t} + 15 t^{3/2} \).

Step 3: Set Up the Integral

The integral to solve is \( F(x) = \int_{1}^{x} \left( 15 \sqrt{t} + 15 t^{3/2} \right) \, dt \). Split this into two separate integrals: \[ F(x) = 15 \int_{1}^{x} \sqrt{t} \, dt + 15 \int_{1}^{x} t^{3/2} \, dt \].

Step 4: Integrate the First Term

For \( 15 \int_{1}^{x} \sqrt{t} \, dt \), rewrite \( \sqrt{t} \) as \( t^{1/2} \). The antiderivative is \[ 15 \left[ \frac{t^{3/2}}{3/2} \right]_{1}^{x} = 15 \times \frac{2}{3}(x^{3/2} - 1^{3/2}) \].This results in \( 10 (x^{3/2} - 1) \).

Step 5: Integrate the Second Term

For \( 15 \int_{1}^{x} t^{3/2} \, dt \), the antiderivative is \[ 15 \left[ \frac{t^{5/2}}{5/2} \right]_{1}^{x} = 15 \times \frac{2}{5}(x^{5/2} - 1^{5/2}) \].This results in \( 6 (x^{5/2} - 1) \).

Step 6: Combine the Results

Add the results of the two integrals together to get the full expression: \[ F(x) = 10 (x^{3/2} - 1) + 6 (x^{5/2} - 1) \] which simplifies to \[ F(x) = 10x^{3/2} + 6x^{5/2} - 16 \].

Key Concepts

Understanding Antiderivatives

An antiderivative represents a function whose derivative gives back the original function you started with. Antiderivatives are essential when dealing with integrals. They help us find the accumulation of quantities, such as area under a curve directly from the function. In our exercise, we aimed to find the antiderivative of the function \( f(t) = 15 \sqrt{t} (1+t) \).
The first step involves rewriting the function into simpler terms like power functions, making it easier to integrate. So, the given function is broken down into two power functions: \( 15 \sqrt{t} + 15 t^{3/2} \). From here, each term is integrated separately.
To integrate, you add 1 to the exponent and divide by the new exponent. This method, known as the power rule of integration, assisted us in finding an antiderivative for each term, leading eventually to the function \( F(x) \). It's about systematically reversing the differentiation process, step by step, to summarize the behavior of the function over an interval.

Using Integration by Parts

While integration by parts wasn't directly used in this exercise, it is worth mentioning as a powerful technique in calculus. Similar to the product rule in differentiation, integration by parts is vital when dealing with products of functions. The general formula is:

• \( \int u \, dv = uv - \int v \, du \)

The choice of \( u \) and \( dv \) is crucial, guiding the strategy for tackling more complex integrals. While this integral problem didn't need it, some similar problems involving multiplication of functions could greatly benefit from this strategic approach. Remember, it's especially helpful when one function is easily differentiable, and another is easily integrable.
Practice identifying when integration by parts is suitable will bolster your integration skills immensely.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the idea of differentiation and integration, showing that they are inverse processes. It has two main parts:

• The first part allows us to find antiderivatives based on the work from definite integrals, emphasizing that integration and differentiation are interconnected.
• The second part helps to evaluate definite integrals if the antiderivative is known, using the expression \( F(b) - F(a) \).

In our exercise, once we found \( F(x) \) by integrating the given function from 1 to \( x \), we could evaluate this integral further using the properties of definite integrals. We substitute the upper and lower bounds to find the change over the interval.
This fundamental connection allows us to compute areas and solve real-world problems involving accumulation more efficiently, linking the concepts seamlessly and creating a clarity between these mathematical frontiers.