Question

Evaluate the integral: \[ \int_0^1 \left( \sqrt{10} \right)^{2x} \, dx \]

• A. \( \frac{10}{\log 10} \)
• B. \( \frac{9}{\log 10} \)
• C. \( \frac{1}{\log 10} \)
• D. \( \frac{4}{\log 5} \)

Hint:

When encountering exponential integrals with a base other than \( e \), use logarithms to simplify the integral.

Answer 1

The Correct Option is B

We are given the integral: \[ I = \int_0^1 \left( \sqrt{10} \right)^{2x} \, dx = \int_0^1 10^x \, dx \] Step 1: Rewrite the integrand as \( 10^x \). Step 2: The integral of \( 10^x \) is: \[ \int 10^x \, dx = \frac{10^x}{\log 10} \] Step 3: Evaluate the integral from 0 to 1: \[ I = \left[ \frac{10^x}{\log 10} \right]_0^1 = \frac{10^1 - 10^0}{\log 10} = \frac{10 - 1}{\log 10} = \frac{9}{\log 10} \] Thus, the correct answer is \( \frac{9}{\log 10} \). % Final Answer The value of the integral is \( \frac{9}{\log 10} \).

Answer 2

Step 1: Write the integral clearly
\[ I = \int_0^1 \left( \sqrt{10} \right)^{2x} \, dx \]

Step 2: Simplify the integrand
Recall that \(\sqrt{10} = 10^{1/2}\).
So, \[ \left( \sqrt{10} \right)^{2x} = \left( 10^{1/2} \right)^{2x} = 10^{x} \].

Step 3: Rewrite the integral
\[ I = \int_0^1 10^{x} \, dx \].

Step 4: Integrate using the exponential formula
Recall the formula:
\[ \int a^{x} \, dx = \frac{a^{x}}{\ln a} + C \], where \(a > 0, a \neq 1\).

Apply it here:
\[ I = \left[ \frac{10^{x}}{\ln 10} \right]_0^1 = \frac{10^{1} - 10^{0}}{\ln 10} = \frac{10 - 1}{\ln 10} = \frac{9}{\ln 10} \].

Final Answer:
\[ \int_0^1 \left( \sqrt{10} \right)^{2x} \, dx = \frac{9}{\log 10} \].
(Note: \(\ln\) and \(\log\) are used interchangeably here as natural logarithm.)