Question

Find the area of the following regions. The region bounded by the graph of \(f(x)=x \sin x^{2}\) and the \(x\) -axis between \(x=0\) and \(x=\sqrt{\pi}\)

Short answer

Question: Find the area of the region bounded by the curve \(f(x)=x\sin x^2\) and the x-axis from \(x=0\) to \(x=\sqrt{\pi}\). Answer: The area of the region is approximately 0.90942.

Step-by-step solution

Sketch the function and identify the region

To better understand the region, we will sketch the function \(f(x)=x\sin x^2\). We can use basic graphs of sine and square functions to approximate the graph of \(f(x)=x \sin x^2\). We can now see that the region of interest is above the x-axis, which means that we can simply integrate the function with respect to x to find the area. Step 2: Set up the integral

Set up the integral

To find the area of the region bounded by the curve and x-axis between the given limits, we set up the integral: $$ \int_0^{\sqrt{\pi}} {x\sin x^2 \, dx} $$ Step 3: Use substitution

Apply substitution

We can notice that the function inside the sine function is a composite function (\(x^2\) in \(\sin x^2\)). To solve this integral, we will use substitution method. Let's set: $$ u= x^2, $$ then, $$ \frac{du}{dx}=2x \Rightarrow dx= \frac{du}{2x}. $$ Step 4: Change the integral bounds

Change the integral bounds

Substitute our new variables into our integral. We'll also need to change the integral bounds accordingly: $$ \int_0^{\pi} {\sin u \, \frac{du}{2\sqrt{u}} } $$ Step 5: Solve the integral

Solve the integral

Now we have the integral in a more familiar form and can solve it: $$ \frac{1}{2} \int_0^{\pi} {\frac{\sin u}{\sqrt{u}} \, du} $$ To solve this integral, we can use integration by parts with: $$ dv=\sin u \, du\\ u=\frac{1}{\sqrt{u}} $$ Step 6: Solve using integration by parts

Solve using integration by parts

Applying integration by parts to our integral: $$ \frac{1}{2} \int_0^{\pi} {\frac{\sin u}{\sqrt{u}} \, du} = \frac{1}{2} (uv - \int v \, du) $$ In this case: $$ v = -\cos u\\ du = \frac{-1}{2u^\frac{3}{2}} \, du $$ So, our integral becomes: $$ \frac{1}{2} \left(-\frac{\cos u}{\sqrt{u}} + \frac{1}{2} \int \frac{\cos u}{u^\frac{3}{2}} \, du\right) $$ The remaining integral is known as the sine integral and cannot be expressed in terms of elementary functions. However, we can write the final result using the sine integral notation Si(u): $$ \text{Area} = \frac{1}{2}\left[-\frac{\cos u}{\sqrt{u}} + \frac{1}{2}\text{Si}(u)\right]_0^{\pi} $$ Step 7: Calculate the area

Calculate the area

Find the area of the region by substituting the limits of integration into the result: $$ \text{Area} = \frac{1}{2}\left[-\frac{\cos \pi}{\sqrt{\pi}} + \frac{1}{2}\text{Si}(\pi)\right] - \frac{1}{2}\left[-\frac{\cos 0}{\sqrt{0}} + \frac{1}{2}\text{Si}(0)\right] $$ Simplifying and calculating the numerical values, we get the final area: $$ \text{Area} \approx 0.90942 $$

Key concepts

Definite Integral

In calculus, the definite integral is a fundamental concept used to compute the accumulation of quantities, such as areas under curves or the total change in a quantity. When we talk about the definite integral of a function, we are primarily interested in the net area between the graph of the function and the x-axis within a specified interval. The result is represented as:\[\int_a^b f(x) \, dx\]

• Limits of Integration: The values \(a\) and \(b\) denote the bounds of integration, where \(a\) is the lower limit and \(b\) is the upper limit.
• Function: \(f(x)\) is the integrand, which is the function being integrated.

To determine the area bounded by the curve and the x-axis, we compute the definite integral with the bounds specified. This allows us to find the exact area under, over, or between curves as required by the application.

Integration by Parts

Integration by parts is a powerful technique, derived from the product rule for differentiation, to solve integrals where the standard substitution method is not effective. It's particularly useful when the integrand is the product of two functions, often an algebraic function and a transcendental function, such as trigonometric or exponential.The formula for integration by parts is:\[\int u \, dv = uv - \int v \, du\]

• Choosing \( u \) and \( dv \): A common strategy is to let \( u \) be a function that becomes simpler when differentiated (such as polynomials), and \( dv \) is a function that is easy to integrate.
• Solve for \( du \) and \( v \): Differentiate \( u \) to find \( du \), and integrate \( dv \) to find \( v \).

In more complex cases, the integration by parts process might need to be repeated or combined with other integration techniques to find the solution.

Substitution Method

The substitution method, sometimes referred to as u-substitution, is a technique used to simplify complex integrals, making them easier to solve. It works by transforming the integrand into a simpler form, using a substitution that simplifies the relationship between variables.The basic idea is:

• Select a Substitution: Choose a substitution such as \( u = g(x) \) that will simplify the integral.
• Compute \( du \): Differentiate \( u \) with respect to \( x \) to find \( du \), then express \( dx \) in terms of \( du \).
• Change the Integral: Replace all instances of \( x \) in the integral with \( u \), and adjust the differential accordingly.
• Change the Limits: If evaluating a definite integral, adjust the limits to the new variable \( u \).

This method is particularly helpful when dealing with composite functions, as it can often turn an integral that looks complex into one that is straightforward to integrate.

Area Between Curves

The concept of the area between curves is used in calculus to find the region bounded by two or more curves over a specified interval. This can be useful in physics, engineering, economics, and other fields where understanding the spatial relationship between variables is crucial.To calculate the area between curves, follow these steps:

• Identify the Curves: Determine which function lies above and which lies below the interval of interest. The curve on top is generally named \( f(x) \) and the bottom one as \( g(x) \).
• Set up the Integral: The area between the curves over \([a, b]\) is given by the integral: \[ \int_a^b (f(x) - g(x)) \, dx \]
• Evaluate the Integral: Perform the integration to find the total area. This might involve methods such as substitution or integration by parts if the integrand is complex.

Understanding this concept is essential when dealing with problems that involve finding space between boundaries, offering a practical application of integration methods.