How does the product of powers law apply to finding the power of a power?

1 answer:

7 0

The product of powers tells us about that multiplying the powers with the same number you have simply have to add the exponents,

for example;

=6^3 * 6^3

=6^3 *6^3 (^3+^3) we are just adding the exponents even thou the operation is multiplication.

= 36^6

The power of power tells us about multiplying both the expression including the exponents. Power rule tells us that a power to a power, just multiply the exponents.

for example;

=6^3 * 6^3

=6^3 * 6^3 (^3*^3) we are multiplying both of the expression

=36^9

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Read 2 more answers

Use the given transformation to evaluate the given integral, where r is the triangular region with vertices (0, 0), (8, 1), and

Jlenok [28]

We first obtain the equation of the lines bounding R.

For the line with points (0, 0) and (8, 1), the equation is given by:

$\frac{y}{x} = \frac{1}{8} \\ \\ \Rightarrow x=8y \\ \\ \Rightarrow8u+v=8(u+8v)=8u+64v \\ \\ \Rightarrow v=0$

For the line with points (0, 0) and (1, 8), the equation is given by:

$\frac{y}{x} = \frac{8}{1} \\ \\ \Rightarrow y=8x \\ \\ \Rightarrow u+8v=8(8u+v)=64u+8v \\ \\ \Rightarrow u=0$

For the line with points (8, 1) and (1, 8), the equation is given by:

$\frac{y-1}{x-8} = \frac{8-1}{1-8} = \frac{7}{-7} =-1 \\ \\ \Rightarrow y-1=-x+8 \\ \\ \Rightarrow y=-x+9 \\ \\ \Rightarrow u+8v=-8u-v+9 \\ \\ \Rightarrow u=1-v$

The Jacobian determinant is given by

$\left|\begin{array}{cc} \frac{\partial x}{\partial u} &\frac{\partial x}{\partial v}\\\frac{\partial y}{\partial u}&\frac{\partial y}{\partial v}\end{array}\right| = \left|\begin{array}{cc} 8 &1\\1&8\end{array}\right| \\ \\ =64-1=63$

The integrand x - 3y is transformed as 8u + v - 3(u + 8v) = 8u + v - 3u - 24v = 5u - 23v

Therefore, the integration is given by:

$63 \int\limits^1_0 \int\limits^{1}_0 {(5u-23v)} \, dudv =63 \int\limits^1_0\left[\frac{5}{2}u^2-23uv\right]^{1}_0 \\ \\ =63\int\limits^1_0(\frac{5}{2}-23v)dv=63\left[\frac{5}{2}v-\frac{23}{2}v^2\right]^1_0=63\left(\frac{5}{2}-\frac{23}{2}\right) \\ \\ =63(-9)=|-576|=576$