Problem 66 - Calculus Single Variable

Question

Integrate by parts. This will result in an integrand of the form \(P(x) / Q(x)\) where \(P(x)\) and \(Q(x)\) are polynomials with the degree of \(P(x)\) greater than or equal to the degree of \(Q(x)\). Such an integrand is handled by performing polynomial division to put \(P(x) / Q(x)\) into the form \(r(x)+s(x) / Q(x)\) where \(r(x)\) and \(s(x)\) are polynomials with the degree of \(s(x)\) less than the degree of \(Q(x)\) $$ \int_{0}^{1} 6 x^{2} \arctan (x) d x $$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{\pi}{2} - \frac{2}{3} \).

Step 1: Identify Parts for Integration by Parts

In the integration by parts formula, \( \int u \ dv = uv - \int v \ du \), we need to choose \( u \) and \( dv \). For the integral \( \int_{0}^{1} 6x^2 \arctan(x) \, dx \), set \( u = \arctan(x) \) (since it's easier to integrate a polynomial form than \( \arctan(x) \)) and \( dv = 6x^2 \, dx \).

Step 2: Differentiate and Integrate Terms

Differentiate \( u = \arctan(x) \) to get \( du = \frac{1}{1+x^2} \, dx \). Integrate \( dv = 6x^2 \, dx \) to get \( v = 2x^3 \).

Step 3: Apply the Integration by Parts Formula

Applying the formula yields: \( \int 6x^2 \arctan(x) \, dx = 2x^3 \arctan(x) \bigg|_{0}^{1} - \int_{0}^{1} 2x^3 \cdot \frac{1}{1+x^2} \, dx \).

Step 4: Evaluate the First Term

For \( 2x^3 \arctan(x) \bigg|_{0}^{1} \), calculate as follows: \( [2(1)^3 \arctan(1) - 2(0)^3 \arctan(0)] = 2 \cdot \frac{\pi}{4} = \frac{\pi}{2} \).

Step 5: Simplify and Integrate the Remaining Term

We need to evaluate \( \int_{0}^{1} \frac{2x^3}{1+x^2} \, dx \). Use polynomial division to express the fraction. Perform the division. If after dividing, it results in a simpler fraction, integrate each part accordingly.

Step 6: Complete the Integration

With polynomial division, find \( r(x) + \frac{s(x)}{1+x^2} \). Integrate both parts: \( \int r(x) \, dx \) and \( \int \frac{s(x)}{1+x^2} \, dx \). Combine these results.

Key Concepts

Polynomial Division

Polynomial division is a method used to simplify complex fractions where the numerator and the denominator are both polynomials. It is similar to the long division process we use in arithmetic, but it's applied to algebraic expressions instead. In the context of integration, polynomial division helps break down a fraction into more manageable parts, making it easier to integrate each component individually.

Here’s how it works:

• Divide the leading term of the numerator by the leading term of the denominator.
• Multiply the entire divisor by this result and subtract it from the original numerator.
• Repeat the process with the remainder until the degree of the remainder is less than the degree of the denominator.

For example, if you have a function in the form \( \frac{2x^3}{1+x^2} \), and you perform polynomial division, you will separate the function into a simpler polynomial plus a remainder over the original divisor, like \( r(x) + \frac{s(x)}{1+x^2} \). This lets you apply different integration techniques to each term, often simplifying the overall problem.

Arctan Function

The arctan function, short for arc tangent, is the inverse of the tangent function. It’s denoted as \( \arctan(x) \), and it converts ratios of sides in right-angled triangles back into angles. Understanding \( \arctan \) is critical when it's part of an integral, as it dictates how differentiation and integration should be carried out.

In integration by parts, setting \( u = \arctan(x) \) is often strategic:

• Derivative: When differentiating, \( \frac{d}{dx} [\arctan(x)] = \frac{1}{1+x^2} \).
• This derivative form is simpler and easier to handle in integration problems.

The arctan function tends to naturally appear in integrals involving trigonometric and algebraic combinations, which often necessitate techniques like integration by parts to solve. This process breaks the integral into more tractable pieces by leveraging the differentiation properties of \( \arctan(x) \).

Definite Integral

A definite integral represents the net area under a curve delineated by a function, usually between two bounds \( a \) and \( b \). In notation, it’s expressed as \( \int_{a}^{b} f(x) \, dx \). This value is a crucial component in calculus, providing insights into accumulated quantities and changes over an interval.

Considerations while evaluating a definite integral:

• Evaluate the antiderivative at the upper limit and subtract the value of the antiderivative at the lower limit.
• Remember that the definite integral reflects signed area, meaning parts of the graph below the x-axis are subtracted from the total.

In the given exercise, such as \( \int_{0}^{1} 6x^2 \arctan(x) \, dx \), the definite integral tells us the total effect of the function, depending on how it behaves across the interval \([0, 1]\). Calculating it involves both applying integration by parts and simplifying results through processes like polynomial division. The final expression represents the complete accumulated area between these bounds, taking into account all simplifying steps undertaken during computations.