91Ó°ÊÓ

Integration by parts is a valuable technique in integral calculus. It is especially useful when dealing with the integral of a product of two functions. The basic formula for integration by parts is given by \[ \int u \, dv = uv - \int v \, du \].To apply this technique, you break down the integral into:

• Identify the parts: Typically, one function is easier to integrate, while the other is easier to differentiate.
• Find the derivatives: For the chosen function \( u \), find \( du \); for \( dv \), find \( v \) by integrating.
• Substitute: Use the formula to replace the original integral with an expression involving \( uv \) and a new integral \( \int v \, du \).

It transforms a difficult integral into simpler parts, making it easier to solve.
In our exercise, we chose \( u = \sin f(x) \) and \( dv = f'(x) \, dx \). This choice meant differentiating \( \sin f(x) \) and integrating \( f'(x) \, dx \), which led to manageable components for substitution.

A definite integral has specified upper and lower limits. It calculates the accumulation of quantities, or area under the curve, between these limits. Represented with notation like this: \[ \int_{a}^{b} f(x) \, dx \], these integrals provide a precise evaluation, unlike indefinite integrals that include a constant of integration. In this context:

• \( a \) and \( b \): The lower and upper limits of integration, respectively.
• The definite integral computes the net signed area under the curve \( f(x) \) between \( x = a \) and \( x = b \).

To evaluate, perform any necessary integration techniques and then use the fundamental theorem of calculus:

• Substitute the upper limit into the antiderivative, then substitute the lower limit.
• Subtract the result of the lower limit from the upper.

In our problem, the definite integral bounds were from 0 to 1, meaning that we computed the result over this interval for \( \int f'(x) \sin f(x) \, dx \). This involved initially integrating by parts before evaluating the boundary expression: \( f(x) \sin f(x) \bigg|_0^1 \).

A differentiable function is a function whose derivative exists at every point in its domain. This ensures that the function is smooth and has no breaks, sharp corners, or vertical tangents. For integrals, it means:

• The function value \( f(x) \) can change smoothly without abrupt interruptions.
• It allows using calculus techniques like differentiation and integration effectively.

The differentiable property is crucial in our task, as:\

• It made it possible to use integration by parts, requiring us to differentiate and integrate \( f(x) \) and \( f'(x) \).

The table values provided in our exercise, both for \( f(x) \) and its derivative \( f'(x) \), enabled detailed evaluation. Since \( f(x) \) was proven differentiable, the use of such calculus operations was justified and reliable.

Evaluating integrals is a key component of integral calculus. It refers to finding the antiderivative and calculating the area under curves as needed. This process becomes specific when constraints, like bound limits for definite integrals, are present.Steps involved generally include:

• Choose the technique: Methods like substitution, partial fractions, or integration by parts ensure accuracy.
• Perform integration: Derive the antiderivative or apply techniques step-by-step.
• Apply limits: In definite integrals, use limits to find specific values.

Returning to our scenario:

• After applying integration by parts, we evaluated expression \( f(x) \sin f(x) \bigg|_0^1 \).
• This was done by substituting values from the given table for \( f(0), f(1), \ldots \).

Ultimately, we combined calculated parts to reach the final evaluation. This method emphasizes correct arithmetic processes and table interpretation for confident resolution of calculus problems.