Question

Calculus Help

Answer 1

Write \[\lim_{n \rightarrow \infty}\sum_{k=1}^{n}(2+k*(5/n))^3 *5/n\] as a definite Integral

Answer 2

Find \[dx/dy \int\limits_{2}^{x^3}\ln(x^2)dx\]

Answer 3

Find the area bounded by the curves y^2= 2x+6 and x=y+1. Work must be an integral with one variable

Answer 4

Write the integral in one variable to find the volume of the solid obtained by rotating the first quadrant region bounded by y=0.5x^2 and y=x about the line x=7.

Answer 5

0.5x² = x
0.5x² - x = 0
x (0.5x - 1) = 0
x = 0, 2
Those are the x-coordinates of the two points of intersection of y = 00.5x² and y=x. The full points are (0,0) and (2,2).

Revolving the region around a vertical axis like x = 3 is better suited to the cylindrical shell method so that you don't have to solve for y. You integrate the lateral surface area of a cylindrical shell having its height vertically within the bounded region. The radius is 3-x (because x is the distance from the y-axis to the outer edge of the shell, leaving you with 3-x as the distance from the outer edge of the shell to x=3). The height of the cylinder is upper curve minus lower curve = x - 0.5x².

Therefore, volume = 2π∫rh dx
= 2π∫ (3-x)(x - 0.5x²) dx, from x = 0 to 2
= 2π∫ (3x-x² - 1.5x² + 0.5x³) dx, from x = 0 to 2
= 2π∫ (3x - 2.5x² + 0.5x³) dx, from x = 0 to 2
= 2π [3x²/2 - 2.5x³/3 + 0.5x⁴/4], from x = 0 to 2
= 2π [6 - 20/3 + 2]
= 2π [4/3] = 8π/3

Answer 6

A ball is thrown upward with a speed of 40 feet per second from the edge of a cliff 500 feet above the ground. What is the speed of the ball when it hits the ground? Use acceleration due to gravity as -32 feet per second squared.

Answer 7

question #2:
the integral is, integrating by parts, is:
\[\Large \begin{gathered}
\int {dx\ln \left( {{x^2}} \right)} = x\ln \left( {{x^2}} \right) - \int {x\frac{{dx}}{{{x^2}}}} 2x = \hfill \\
\hfill \\
= x\ln \left( {{x^2}} \right) - 2x \hfill \\
\end{gathered} \]

Answer 8

so the definite integral is:
\[\Large \left. {x\ln \left( {{x^2}} \right) - 2x} \right|_2^{{x^3}} = {x^3}\ln \left( {{x^6}} \right) - 2{x^3} - 2\ln 4 + 4\]

Answer 9

for iluvhomewurk's did they just confuse 3 with 7?

Answer 10

Question #3
we can rewrite your equations as follows:
\[\Large \left\{ \begin{gathered}
x = \frac{{{y^2} - 6}}{2} \hfill \\
x = y + 1 \hfill \\
\end{gathered} \right.\]

Answer 11

the intersection point are:
\[\begin{gathered}
P = \left( { - 1, - 2} \right) \hfill \\
Q = \left( {5,4} \right) \hfill \\
\end{gathered} \]

Answer 12

so the requested area, is given by the subsequent integral:
\[\Large \int_{ - 2}^4 {dy\left( {y + 1 - \frac{{{y^2} - 6}}{2}} \right)} \]

Answer 13

are you still talking about question 3?

Answer 14

yes!

Answer 15

here is the next step:
\[\Large \int_{ - 2}^4 {dy\left( {y + 1 - \frac{{{y^2} - 6}}{2}} \right)} = \left. {y - \frac{{{y^2}}}{2} + 4} \right|_{ - 2}^4 = 28\]