Question

Evaluate the partial integral. $$ \int_{0}^{x}(2 x-y) d y $$

Short answer

The result of the evaluation of the partial integral is \( \frac{3}{2}x^2 \)

Step-by-step solution

Integrate the function

The first step is to integrate the function \(2x - y\) with respect to y over the interval [0, x]. The integral of this function can be obtained by treating x as a constant because we are integrating with respect to y. Hence, the first term \(2x \times y\) becomes \(2xy\), and the second term \(-y\) integrates to \(-\frac{1}{2}y^2\). So, the integral becomes: \[ \int (2x - y) dy = 2xy - \frac{1}{2}y^2 \]

Apply the limits of the integration

Substitute the limits of the integral into the equation. We substitute \(y = x\) in place of \(y\) for the upper limit and \(y = 0\) for the lower limit. The integral then becomes: \[ 2x^2 - \frac{1}{2}x^2 - (2x*0 - \frac{1}{2}*0^2) = 2x^2 - \frac{1}{2}x^2 \]

Simplify the result

Lastly, simplify the result from Step 2 to bring it to its simplest form. The difference of \(2x^2\) and \(\frac{1}{2}x^2\) results in \(\frac{3}{2}x^2\). So the final result of the integral is: \[ \frac{3}{2}x^2 \]

Key concepts

Definite Integral

A definite integral is an essential tool in calculus, acting like a summation operator that aggregates small quantities. It evaluates the area under a curve, giving us insight into the total accumulation of a function between two bounds. Here, we are dealing with the integral from 0 to x for the function \(2x - y\). To solve it, we find the antiderivative (essentially the reverse of differentiation) first, and then calculate using the given limits.

• Definite Integrals have two limits: the lower and upper limits. In our example, these limits are 0 and x.
• The result provides the accumulated value of the function \(2x-y\) over the interval from y=0 to y=x.

Understanding definite integrals gives us a method to handle real-life scenarios like calculating total distance, area, or any cumulative measure.

Integration by Substitution

Integration by substitution, often called 'u-substitution,' is a technique to simplify the integration process. This method allows us to transform the integral into a simpler form by changing variables. However, in this particular example, a simpler method suffices because the variable \(x\) is treated as a constant during integration over \(y\).

• When substitution is needed, you often replace a complex portion with a single variable to make integration easier.
• The aim is clear, simplify the integrand (the expression you're integrating) for easier calculation.

Although it's not directly applied in this exercise, knowing when to use substitution is a noteworthy skill in tackling more complex integrals.

Calculus Techniques

Calculus comes with an arsenal of techniques to solve integrals, each applicable to different types of functions and problems. Partial integration, which we are using here, is perfect when integrating expressions with constants and simple variables.

• Step-by-step approach: Start by identifying what you are integrating with respect to and treat other variables as constants.
• Careful treatment of variables: Recognize that treating variables appropriately is key. Here, with respect to \(y\), \(x\) becomes a constant more so like a coefficient.
• Simplify carefully: Always recheck whether subtraction or addition is required after applying the limits.

Familiarity with different calculus techniques is vital as it allows you to choose the most efficient solution method for a given problem.

Mathematical Problem Solving

Mathematics is not just about applying formulas; it is also an exercise in logical thinking and problem-solving. By understanding the entire process from setting up the integral correctly to simplifying it at the end, you become more adept at handling various mathematical challenges.

• Start with the problem's breaking down: Decompose the problem into smaller parts and solve each part systematically.
• Verify each step: Double-check each mathematical operation to ensure accuracy.
• Practice interpretation: Be sure to understand what the final result represents in context with the original problem.

As you gain more practice with calculus and integrals, your ability to solve complex mathematical problems increases, making seemingly tough exercises become manageable.