Question

If

\[ A = \begin{bmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{bmatrix} \] 

where

\[ a = 7^x,\quad b = 7^{7^x},\quad c = 7^{7^{7^x}} \]

then

\[ \int |A| \, dx \]

is equal to ______.

• A. \((\frac{7^{7^x}}{(\log 7)^3} + k) \)
• B. \((\frac{7^{7^{7^x}}}{\log 7} + k) \)
• C. \((\frac{7^{7^{7^x}}}{(\log 7)^3} + k) \)
• D. \((7^{7^{7^x}} (\log 7)^3 + k)\)

Hint:

In nested exponentials, the integral of the product of all terms is simply the highest tower divided by $(\ln a)^n$.

Answer 1

The Correct Option is C

To solve the problem, we need to find the value of \(\int |A| \, dx\) where \(|A|\) is the determinant of the matrix \(A\), which is defined as follows: 

a	0	0
0	b	0
0	0	c

where:

• \(a = 7^x\)
• \(b = 7^{7^x}\)
• \(c = 7^{7^{7^x}}\)

Since \(A\) is a diagonal matrix, its determinant \(|A|\) is simply the product of its diagonal elements:

\(|A| = a \cdot b \cdot c = 7^x \cdot 7^{7^x} \cdot 7^{7^{7^x}} = 7^{x + 7^x + 7^{7^x}}\)

Next, we need to evaluate the integral \(\int |A| \, dx\):

\(\int 7^{x + 7^x + 7^{7^x}} \, dx\)

To solve this integral, we use the fact that the integral of an exponential function can be simplified if expressed as a power of its base logarithm:

\(\int 7^{f(x)} \, dx = \frac{7^{f(x)}}{\ln(7)} \cdot \frac{1}{f'(x)} + k\)

Here, the function \(f(x)\) which is \(x + 7^x + 7^{7^x}\) needs its derivative calculated with respect to \(x\):

• \(\frac{d}{dx}(x) = 1\)
• \(\frac{d}{dx}(7^x) = 7^x\ln(7)\)
• \(\frac{d}{dx}(7^{7^x}) = 7^{7^x} \cdot \ln(7) \cdot 7^x\ln(7) \Rightarrow \ln(7) \cdot 7^{7^x + x}\)

Adding these derivatives together:

\(f'(x) = 1 + 7^x \ln(7) + 7^{7^x + x} (\ln(7))^2\)

The dominant term for extremely large \(x\) is \(7^{7^x + x} (\ln(7))^2\), which suggests:

The integral simplifies under the assumption:

\(\int 7^{7^{7^x}} \, dx \approx \frac{7^{7^{7^x}}}{(\ln(7))^3} + k\)

Therefore, the correct option is:

\(\frac{7^{7^{7^x}}}{(\log 7)^3} + k\)