Problem 2
Question
Evaluate the partial integral. $$ \int_{x}^{x^{2}} \frac{y}{x} d y $$
Step-by-Step Solution
Verified Answer
The value of the given definite integral is \(\frac{x^{3}}{2} - \frac{x}{2}\)
1Step 1: Define the Integral Function
First, realize the problem is to evaluate the integral \(\int_{x}^{x^{2}} \frac{y}{x} dy\). The function to be integrated is \(f(y) = \frac{y}{x}\), and the limits of the integral are \(x\) (lower limit) and \(x^{2}\) (upper limit). Note that 'x' is considered as a constant with respect to 'y' as our integral is with respect to 'y'.
2Step 2: Integration Process
Now, integrate function \(f(y) = \frac{y}{x}\). This becomes \(\int \frac{y}{x} dy\). Since 'x' can be considered constant in terms of 'y', the integral becomes \(\frac{1}{x} \int y dy\), which after integrating is \(\frac{y^{2}}{2x}\).
3Step 3: Application of Limits
Next, apply the limits to the integral, from 'x' to \(x^{2}\). This becomes \(\frac{1}{2x}[ (x^{2})^{2} - x^{2} ] =\frac{x^{4} - x^{2}}{2x}=\frac{x^{3}}{2} - \frac{x}{2}\).
4Step 4: Simplify
Finally, simply the result to get the answer. In this case, we don't need to simplify it anymore.
Other exercises in this chapter
Problem 1
Plot the points on the same threedimensional coordinate system. (a) \((2,1,3)\) (b) \((-1,2,1)\)
View solution Problem 2
Sketch the region of integration and evaluate the double integral. $$ \int_{0}^{3} \int_{0}^{1}(2 x+6 y) d y d x $$
View solution Problem 2
Use Lagrange multipliers to find the given extremum. In each case, assume that \(x\) and \(y\) are positive. $$ \text { Maximize } f(x, y)=x y \quad 2 x+y=4 $$
View solution Problem 2
Find any critical points and relative extrema of the function. $$ f(x, y)=x^{2}+y^{2}+2 x-6 y+6 $$
View solution