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 the value of \[ \int |A| \, dx \] is:

• 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

Step 1: Understanding the Question:
The question asks to find the integral of the determinant of a diagonal matrix $A$ whose diagonal elements are exponential functions in terms of $x$.
Step 2: Key Formula or Approach:
For a diagonal matrix, the determinant $|A|$ is the product of its diagonal elements.
$|A| = a \cdot b \cdot c = 7^x \cdot 7^{7^x} \cdot 7^{7^{7^x}}$.
We will use substitution method for integration.
Step 3: Detailed Explanation:
We need to calculate $I = \int 7^{7^{7^x}} \cdot 7^{7^x} \cdot 7^x dx$.
Let $u = 7^{7^{7^x}}$.
Differentiating with respect to $x$:
$\frac{du}{dx} = 7^{7^{7^x}} \cdot \log 7 \cdot \frac{d}{dx} (7^{7^x})$
$\frac{du}{dx} = 7^{7^{7^x}} \cdot \log 7 \cdot (7^{7^x} \cdot \log 7 \cdot \frac{d}{dx} (7^x))$
$\frac{du}{dx} = 7^{7^{7^x}} \cdot \log 7 \cdot 7^{7^x} \cdot \log 7 \cdot (7^x \cdot \log 7)$
$\frac{du}{dx} = (7^{7^{7^x}} \cdot 7^{7^x} \cdot 7^x) \cdot (\log 7)^3$
$\implies (7^{7^{7^x}} \cdot 7^{7^x} \cdot 7^x) dx = \frac{du}{(\log 7)^3}$
Now substitute this in the integral:
$I = \int \frac{1}{(\log 7)^3} du = \frac{u}{(\log 7)^3} + k$
Substituting $u$ back:
$I = \frac{7^{7^{7^x}}}{(\log 7)^3} + k$
Step 4: Final Answer:
The integral is $\frac{7^{7^{7^x}}}{(\log 7)^3} + k$.