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Chapter 7: Problem 1

In Exercises \(1-10,\) evaluate the partial integral. $$ \int_{0}^{x}(2 x-y) d y $$

Short Answer

Expert verified
The integral is \( \frac{3}{2}x^2 \).

Step by step solution

01

Identify the Limits of Integration

The limits of integration are given as 0 and x. This means the integral will be evaluated at these points.
02

Apply the Integral

To apply the integral to the function, we integrate the function inside the brackets with respect to y: \( \int (2x-y) dy \). Applying the integral yields \(2xy - \frac{y^2}{2}\).
03

Evaluate the Integral

Now, evaluate the function at the upper and lower limits of integration, which are x and 0 respectively. This gives: \( [2x*x - \frac{x^2}{2}] - [2x*0 - \frac{0^2}{2}] \) which simplifies to: \( 2x^2 - \frac{x^2}{2} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

limits of integration
When performing integrals, the limits of integration define the interval over which we evaluate the integral. In this exercise, the limits are given as 0 and \(x\). This tells us where we start and stop measuring the area under the curve. The lower limit is 0, and the upper limit is \(x\). Both of these points are crucial for solving the integral.
To properly evaluate the integral, you substitute these limits into the result of the integration. This results in two "endpoint" evaluations, typically represented as \(F(b) - F(a)\), where \(b\) and \(a\) are the upper and lower limits, respectively.
integral evaluation
Evaluating an integral involves finding the antiderivative and then calculating its value at the specified limits of integration. In this problem, after integrating \((2x - y)\) with respect to \(y\), we get the expression \(2xy - \frac{y^2}{2}\).
To evaluate this integral, we substitute the limits of integration into this expression: First plug in the upper limit \(x\), yielding \(2x*x - \frac{x^2}{2}\). Then plug in the lower limit 0, which simplifies everything to zero.
Finally, compute the difference between these two results. This gives us the value of the integral over the specified interval.
integrating functions
Integrating a function involves finding the antiderivative of that function with respect to the indicated variable. Here, the task is to integrate \(2x - y\) with respect to \(y\).
The function pieces are independent of \(y\), except for the term \(-y\). So, using basic integration rules:
  • The integral of \(2x\) with respect to \(y\) is simply \(2xy\) because \(2x\) is treated as a constant.
  • The integral of \(-y\) is \(-\frac{y^2}{2}\).
Together, these give the integrated function: \(2xy - \frac{y^2}{2}\).
calculus problem solving
Solving calculus problems often involves step-by-step processes to transform an expression or problem into a solution. For this problem, breaking down the task into manageable parts is essential.
  • First, determine the limits of integration to know where to evaluate the result.
  • Next, integrate the function by finding its antiderivative.
  • Finally, substitute the limits of integration into the antiderivative and solve the resulting expression.
Each of these steps requires a foundational understanding of calculus concepts but can be tackled one at a time to simplify the process. In this way, complex calculations become achievable by focusing on systematic progression through each analytical step.

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Most popular questions from this chapter

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