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a. Write the formula for \(\int f(x) d x\). b. Write the formula for \(\frac{d}{d x} \int f(x) d x\). $$ f(x)=15 \sqrt{x} $$

Short Answer

Expert verified
a) \(\int 15\sqrt{x} \, dx = 10x^{3/2} + C\) b) \(\frac{d}{dx} = 15\sqrt{x}\)

Step by step solution

01

Setting up the integral

We start by setting up the integral \[\int f(x) \, dx\]with the given function \(f(x) = 15 \sqrt{x}\). This translates to \[\int 15 \sqrt{x} \, dx\]
02

Simplify the function inside the integral

Rewrite the square root \(\sqrt{x}\) in terms of exponents: \(x^{1/2}\). Thus, the integral becomes\[\int 15x^{1/2} \, dx\]
03

Apply the power rule for integration

Use the power rule for integration: \[\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\]where \(C\) is the constant of integration. For the integral \(\int 15x^{1/2} \, dx\), we have \(n = 1/2\). So the integral becomes\[15 \cdot \frac{x^{1/2+1}}{1/2+1} + C = 15 \cdot \frac{x^{3/2}}{3/2} + C\]
04

Simplify the integrated expression

Simplify the expression by multiplying:\[15 \cdot \frac{2}{3}x^{3/2} + C = 10x^{3/2} + C\]Thus, the integral is\[\int 15 \sqrt{x} \, dx = 10x^{3/2} + C\]
05

Differentiate the integrated function

To find \(\frac{d}{dx} \int f(x) \, dx\), differentiate \(10x^{3/2} + C\). Differentiating term by term, we find:\[\frac{d}{dx} \left(10x^{3/2} + C \right) = 10 \cdot \frac{d}{dx} \left(x^{3/2}\right) = 10 \cdot \frac{3}{2}x^{1/2} = 15\sqrt{x}\]
06

Confirm the derivative result

The derivative of the integral function is\[15 \sqrt{x}\]which confirms that\[\frac{d}{dx} \int f(x) \, dx = f(x)\]as expected by the Fundamental Theorem of Calculus.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus is a key concept connecting differentiation and integration, two core operations in calculus. This theorem states that if a function is continuous over an interval, then the integral of that function from a point "a" to "x" is a function whose derivative is the original function. Essentially, it provides a way to reverse differentiation. It's two-fold:
  • First Part: If you have a function that is integrable in an interval [a, b], the integral of the function from a to x, is an antiderivative of the function.
  • Second Part: If you find an antiderivative of a function over an interval, evaluating that antiderivative at specific points gives you the value of the definite integral.
In simpler terms, integrating a rate of change (like velocity) across a range gives you the total change (like distance), and differentiating the integral brings you back to the original function. This theorem is what let's us conclude that after integrating and then differentiating the function from the exercise, we retrieve the original function, just as expected.
Power Rule for Integration
The Power Rule for Integration is an essential tool in calculus that simplifies the process of finding integrals for functions of the form \(x^n\), where "n" is any real number except -1. The rule states:
  • For \( \int x^n \, dx \), the integral is \( \frac{x^{n+1}}{n+1} + C \).
The constant "C" represents the constant of integration, accounting for any constant term that could have been present in the original function. This rule follows directly from the reverse process of differentiation, targeting monomial expressions such as \(x^{1/2}\).

When applying to a specific exercise, say \( \int 15 \sqrt{x} \, dx \), which can be rewritten to \(15x^{1/2} \), you use the Power Rule to integrate the expression. This involves incrementing the exponent by 1 and dividing by the new exponent, simplifying it further. So with our previous setup, \( \int 15 \sqrt{x} \, dx \) becomes \( 10x^{3/2} + C \).
The Power Rule for Integration is a straightforward yet powerful rule that assists in simplifying and solving integrals efficiently.
Differentiation
Differentiation is the process of finding a derivative, which represents the rate of change or slope of a curve. In the context of the power of integration, differentiation can be used to check the correctness of an integral calculation.

For instance, if you integrate a function and then differentiate the result, you should obtain the function you started with, assuming the integration was indefinite. This is because of the inverse relationship inherent in the calculus processes of integration and differentiation. The formula \( \frac{d}{dx} [\int f(x) \, dx] = f(x) \) shows this reciprocal nature.

In the exercise, we saw that starting with the integral \( \int 15 \sqrt{x} \, dx \), a derivative check was performed by differentiating \( 10x^{3/2} + C \). This step reaffirmed our solution, demonstrating differentiation by bringing us back to the function \( 15 \sqrt{x} \). Differentiation not only helps identify underlying patterns of change but also confirms accurate integrals, underscoring its vital role in calculus.

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