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Chapter 5: Problem 27

Find the area below the graph of \(f\). $$f(x)=x \sqrt{2 x^{2}+1} , \quad x \in[0.2]$$

Short Answer

Expert verified
The area under the curve of the function \(f(x)=x \sqrt{2 x^{2}+1}\) from \(x=0\) to \(x=2\) is \(\frac{1}{2} 3\sqrt{3}-\frac{1}{12}\).

Step by step solution

01

Understand the problem

The function \(f(x)=x \sqrt{2 x^{2}+1}\) is defined on the interval from 0 to 2. We need to find the area under the curve of this function within this range, which can be calculated using definite integration.
02

Setup the integral

We setup the definite integral of the function to find area under the curve. This integral extends from 0 to 2, which is the range of \(x\). The integral looks like this: \(\int_{0}^{2} x \sqrt{2 x^{2}+1} dx\).
03

Solve the integral

The given integral seems a bit tricky at first, but if we apply a simple substitution, it becomes quite straightforward. Let's set \(u=2x^2+1\), then \(du=4x dx\) or \(dx=du/4x\). Substituting these into our integral, we get: \(\int_{u(0)}^{u(2)} 0.25 \sqrt{u} du\) where \(u(0)=1\) and \(u(2)=9\). Next, we solve the integral: \(\int_{1}^{9} 0.25 \sqrt{u} du = \left[1/3*0.25*u^{3/2}\right]_{1}^{9} = 1/3*0.25*(9^{3/2}-1^{3/2}) = 1/2*3\sqrt{3}-1/12\).
04

Interpret the result

The value of the integral we've just calculated therefore represents the area under the curve of the function from \(x=0\) to \(x=2\). This is the answer to our problem.

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Key Concepts

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

Area Under the Curve
When working with functions in calculus, one common goal is to find the "area under the curve." This concept refers to the region between the graph of a function and the x-axis over a specified interval.
For instance, in the problem involving the function \(f(x) = x \sqrt{2x^2 + 1}\), our task was to calculate this area from \(x = 0\) to \(x = 2\). The area under the curve symbolizes the accumulation of values of the function over the interval.
  • This area is particularly useful in physics and engineering to understand quantities like distance, probabilities, and mass.
  • It is often calculated using definite integrals in calculus, providing a numerical value that represents the space under the curve.
By applying the right integration techniques, we can find this area accurately.
Substitution Method
The substitution method is a valuable technique in integral calculus, often used to simplify complex integrals. In our exercise, we employed this method to integrate the function \(f(x) = x \sqrt{2x^2 + 1}\).
This approach involves replacing a part of the integral with a single variable to make it easier to work with. We set \(u = 2x^2 + 1\) and derived \(du = 4x \, dx\), simplifying the problem.
  • The substitution transforms the original integral into one that's more manageable. In this case, it became \(\int_{1}^{9} 0.25 \sqrt{u} \, du\), which is easier to solve.
  • Remember to change the limits of integration to match the new variable \(u\), resulting in a different interval.
This method efficiently simplifies otherwise complicated integration tasks, enabling key calculations in calculus and beyond.
Integral Calculus
Integral calculus is a vital branch of mathematics focused on the concepts of integration and accumulation. It's about finding the sum of infinitely many infinitesimals to determine total quantities, like areas and volumes.
In our exercise, we used integral calculus to find the area under the curve of a function on a given interval. Definite integrals help us compute these areas with precise mathematical tools.
  • The integral \(\int_{0}^{2} x \sqrt{2x^2+1} \, dx\) is an example of using integral calculus to solve real problems.
  • This calculus branch complements differential calculus, which deals with rates of change, making them powerful together.
Integral calculus provides essential insights into real-world applications across various scientific disciplines.

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