I realized that I reduced to upper triangular row echelon form rather than just upper triangular form.
The next part asks me to check this determinant by expanding the third row. How do I go about expanding it? Would it be using cofactors etc.?
The determinant would be 1.
However, the answer is listed as -12, and the second part is to check my answer by expanding the third row of the determinant. I'm at a loss as to what to do now.
Use row and/or column operations to simplify the determinant of the following matrix A, by reduction to upper triangular form, then evaluate.
A = \left(\begin{array}{cccc}
2 & 3 & 4 & 5\\
0 & -1 & 2 & 1\\
0 & 0 & 2 & 4\\
0 & 3 & -6 & 0
\end{array}
\right)
Is there an simpler way to...
The augmented matrix of a system of linear equations in the unknowns x, y, z, u and v, has the form
\left(\begin{array}{ccccc|c}
1 & -3 & 1 & -1 & 0 & -1\\
0 & 0 & 1 & 1 & 2 & 1\\
0 & 0 & 0 & 0 & 1 & -1\\
0 & 0 & 0 & 0 & 0 & 0
\end{array}
\right)
Write down all solutions of the system...
Alright, I did that and I get V = 2\pi \left[\frac{2}{3}x^3 - \frac{1}{4}x^4 - \frac{1}{2}x^2\right]_{0}^{1}
Then when I put in 1 and 0, my final volume is a negative value...
Oh, sorry about that typo. If revolving around the y-axis it would be:
\pi \int_0^1 (y)^2 - (\sqrt{1 - y} +1)^2 dy.
So the answer when integrating with respect to y and revolved around x = 1 would be the same if I integrated with respect to x and replaced x with x-1?
Find the volume of the solid generated by revolving the region bounded by y = 2x - x^2 and y = x about the line x = 1
Ok, so I can solve this if it were revolved around the y-axis, it would be: \pi \int_0^1 (x)^2 - (2x - x^2)^2 dy
If I have to revolve it around x = 1, what do I need to change...
Yup, I did that and I got I_n = \frac{1}{2}x^ne^{2x} - \frac{n}{2}I_{n-1}
For the second question, I let u = ln x, but what would I let dv/dx equal to?
Hey, I got a couple of problems on integration that I can't seem to figure out.
1. Suppose I_n = \int_{0}^{1} x^n e^{2x} dx Evaluate I_n in terms of I_{n-1} for any natural number n
2. Suppose I_n = \int_{1}^{e} \left[\ln x\right]^n dx Evaluate I_n in terms of I_{n-1} for any natural...
Ok, I think I got you. I chose to integrate with respect to x first.
\int_{0}^{4}\int_{-\sqrt{4-y}+3}^{3} 2xy\,dx\,dy
\int_{0}^{4}[x^2 y]_{-\sqrt{4-y}+3}^{3}\,dy
[5y^2 - 7y - 4(4 - y)^\frac{3}{2}]_{0}^{4}
Answer is 52 units cubed.
x = \sqrt{4 - y} + 3
x = \pm2 + 3 = 1\,\mbox{or}\,5
As x \leq 3 choose x = 1.
After solving the integral, I get:
\int_{1}^{3}\int_{0}^{4} 2xy\,dy\,dx = 64