Question

If (A = a & 0 & 0
0 & b & 0
0 & 0 & c ) where (a = 7^x, b = 7^{7^x}, 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

Step 1: Find determinant (|A|)
For a diagonal matrix, (|A| = a \cdot b \cdot c = 7^x \cdot 7^{7^x} \cdot 7^{7^{7^x}}).

Step 2: Recognize the derivative pattern
Let (u = 7^{7^{7^x}}).
(\frac{du}{dx} = 7^{7^{7^x}} \cdot \ln 7 \cdot \frac{d}{dx}(7^{7^x}) = 7^{7^{7^x}} \cdot \ln 7 \cdot 7^{7^x} \cdot \ln 7 \cdot 7^x \cdot \ln 7).
(\frac{du}{dx} = (7^{7^{7^x}} \cdot 7^{7^x} \cdot 7^x) \cdot (\ln 7)^3).

Step 3: Integrate
(\int (7^{7^{7^x}} \cdot 7^{7^x} \cdot 7^x) , dx = \frac{u}{(\ln 7)^3} + k = \frac{7^{7^{7^x}}}{(\ln 7)^3} + k).
Final Answer: (C)