Problem 31 - Calculus Early Transcendentals

Question

Evaluate the following definite integrals. $$\int_{0}^{\pi} x \sin x d x$$

Step-by-Step Solution

Answer

Answer: The value of the given definite integral is $$\pi$$.

Step 1: Choose u and dv

In this problem, let's choose $$u = x$$ and $$dv = \sin x \, dx$$. This choice is made to have a simpler function to evaluate after deriving u. Step 2: Compute du and v

Step 2: Find du and v

Now we need to find $$du$$ and $$v$$. Differentiate $$u = x$$ to get $$du = dx$$ and integrate $$dv = \sin x \, dx$$ to get $$v = -\cos x$$. Step 3: Integration by Parts formula

Step 3: Apply integration by parts

Using the integration by parts formula, we have: $$\int x \sin x \, dx = -x\cos x - \int -\cos x \, dx$$ Step 4: Simplify the remaining integral

Step 4: Integrate

Now, integrate the remaining integrand function: $$\int \cos x \, dx = \sin x$$ Step 5: Combine the results

Step 5: Obtain the antiderivative

Combine the integration by parts result with the last integral result: $$\int x \sin x \, dx = -x\cos x + \sin x + C$$ where C is the constant of integration. Step 6: Evaluate the definite integral

Step 6: Apply the Fundamental Theorem of Calculus

Since we are given the limits of integration, we can now apply the Fundamental Theorem of Calculus. The definite integral of $$x\sin x$$ from 0 to $$\pi$$ is: $$\int_{0}^{\pi} x \sin x d x = \left[-x\cos x + \sin x\right]_0^{\pi}$$ Step 7: Substitute the limits

Step 7: Evaluate at the limits

Now, evaluate the antiderivative at the limits of integration: $$= -\pi\cos \pi + \sin \pi - (-0\cos 0 + \sin 0) = 0 - 0 + \pi - 0 = \pi$$ Step 8: Final answer

Step 8: Write the answer

Therefore, the value of the given definite integral is: $$\int_{0}^{\pi} x \sin x d x = \pi$$

Key Concepts

Integration by Parts

Learning about integration by parts is akin to having a nifty tool in your calculus toolkit—it simplifies problems that at first glance seem complicated. Take the integral \( \int x \sin x \, dx \), for example. In this technique, we pick two functions within our integral that we'll label as 'u' (typically the one that gets simpler when derived) and 'dv' (usually the one that's manageable when integrated). Here, we chose \( u = x \) and \( dv = \sin x \, dx \).

After we differentiate \( u \) to get \( du \) and integrate \( dv \) to obtain \( v \) (finding \( du = dx \) and \( v = -\cos x \) respectively), we're set to apply the integration by parts formula. This magical formula is \( \int u \, dv = uv - \int v \, du \), which elegantly splits the integral into two parts—one straightforward multiplication and another simpler integral to solve.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is the bridge that connects differentiation and integration. It is a pivotal concept in understanding why we do what we do when finding the area under a curve or the accumulation of quantities. The theorem has two parts, but here we're most interested in the part that says once we find an antiderivative of a function, we can evaluate the definite integral between two points.

What this means for our problem—the definite integral from 0 to \( \pi \) of \( x\sin x \)—is that after we diligently work through finding the antiderivative, we get to use the theorem to plug in the upper and lower bounds of our integral. Subtracting the value of the antiderivative at the lower bound from its value at the upper bound gives us the exact area of the region bounded by the function, the x-axis, and the vertical lines at \( x=0 \) and \( x=\pi \)—which in this problem turns out to be \( \pi \).

Antiderivative

An antiderivative, also known as an indefinite integral, is essentially a function that 'reverses' differentiation. If you have a function \( f(x) \), its antiderivative is another function \( F(x) \) such that \( F'(x) = f(x) \). Why is this important? Because finding the antiderivative is a necessary step in computing a definite integral. The process involves a bit of detective work, where our clues are the differentiation rules we've learned.

In our problem, we find that the antiderivative of \( x\sin x \) is not straightforward and requires integration by parts to work out. By unraveling the problem step by step, we end up with \( -x\cos x + \sin x \) plus a constant \( C \)—this is our antiderivative. Integrals are like stories, and antiderivatives are the plot twists that bring resolution; they tell us what accumulation takes place from one point to another, shedding light on the overall narrative of the function's behavior.

U-Substitution

Imagine if changing clothes could turn complicated tasks into simpler ones—that's u-substitution in the realm of calculus. While not used in our specific problem, u-substitution, also known as variable substitution, is an essential trick when dealing with integrals that aren't so cooperative. It involves choosing a part of the integrand to be \( u \), which simplifies the integral by transforming it into a new variable and hopefully, into a more familiar form.

When we perform u-substitution, we're essentially reversing the chain rule of differentiation. Calculus is really about patterns, and recognizing these patterns allows us to shift perspectives and solve problems more efficiently. Once we've made our substitution and calculated the integral with respect to \( u \), we substitute back into the original variable to complete the solution. Although it isn't required in our current example, it's a technique worth mastering for its problem-solving power in other integral challenges.