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Chapter 12: Problem 1004

The value of the integral \({ }^{1} \int_{0} 2^{2 \mathrm{x}} \cdot 3^{-\mathrm{x}} \mathrm{dx}\) is \(\ldots \ldots \ldots\) (a) \(\log _{\mathrm{e}}(64 / 27)\) (b) \(\log _{\mathrm{e}}(27 / 64)\) (c) \(\log _{(3 / 4)}\) e (d) \(\log _{(64 / 27)} \mathrm{e}\)

Short Answer

Expert verified
The value of the integral is \(\boxed{\log_\mathrm{e}\left(\frac{64}{27}\right)}\).

Step by step solution

01

Rewrite the integral expression

In order to simplify the integral, we rewrite the expression inside the integral using exponent properties: \(2^{2x}\cdot 3^{-x} = (2^2)^x \cdot 3^{-x} = (4)^x \cdot 3^{-x} = \left(\frac{4}{3}\right)^x\) Now, our integral becomes: \(\int_{0}^{1} \left(\frac{4}{3}\right)^x dx\)
02

Evaluate the integral

Next, we evaluate the integral using the known formula for integrating exponentials: If \(\int a^x dx = \frac{a^x}{\ln a} + C\), then: \(\int_{0}^{1} \left(\frac{4}{3}\right)^x dx = \frac{1}{\ln\left(\frac{4}{3}\right)}\left[\left(\frac{4}{3}\right)^x\right]_0^1 = \frac{\left(\frac{4}{3}\right) - 1}{\ln\left(\frac{4}{3}\right)}\)
03

Simplify the expression

Now, we simplify the expression: \(\frac{\left(\frac{4}{3}\right) - 1}{\ln\left(\frac{4}{3}\right)} = \frac{\left(\frac{4}{3}\right) - \frac{3}{3}}{\ln\left(\frac{4}{3}\right)} = \frac{\frac{1}{3}}{\ln\left(\frac{4}{3}\right)} = \frac{1}{3\ln\left(\frac{4}{3}\right)} = \frac{1}{\ln\left(\frac{4}{3}\right)^3}\)
04

Match result with an option

Now, we match our result with one of the given options: Our result \(\frac{1}{\ln\left(\frac{4}{3}\right)^3}\) matches option (a): \(\log_\mathrm{e}\left(\frac{64}{27}\right)\), as the natural logarithm \(\ln\) is the same as \(\log_\mathrm{e}\). Therefore, the value of the integral is \(\boxed{\log_\mathrm{e}\left(\frac{64}{27}\right)}\).

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Most popular questions from this chapter

\(1 \int_{0} 3 \sqrt{\left(x^{3}-x^{4}\right) d x}\) is equal to....... (a) \((1 / 2)\) (b) \((3 / 7)\) (c) \((9 / 28)\) (d) \((29 / 28)\)

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