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Chapter 11: Problem 898

If \(\int\left[\left(2^{\\{1 /(x) 2\\}}\right) \mathrm{d} x\right]=k 2^{\\{1 /(x) 2\\}}+c\) then \(k=\) (a) \(-[1 /(2 \log 2)]\) (b) \(-\log 2\) (c) \(-2\) (d) \(-(1 / 2)\)

Short Answer

Expert verified
The value of \(k\) is \(-\frac{1}{2\log{2}}\). Thus, option (a) is correct.

Step by step solution

01

Rewrite the given equation

We are given the integral of the function \( 2^\frac{1}{2x} \) and its result. The goal is to differentiate this integral with respect to x and find the value of k. Rewrite the given integral equation as: \[ \int \left( 2^{\frac{1}{2x}} \right) \mathrm{d}x = k \, 2^{\frac{1}{2x}} + c \]
02

Differentiate both sides of the equation

To find the value of k, we will differentiate both sides of the equation with respect to x: \[ \frac{d}{dx} \left( \int \left( 2^{\frac{1}{2x}} \right) \mathrm{d}x \right) = \frac{d}{dx} \left( k \, 2^{\frac{1}{2x} } + c \right) \]
03

Use the Fundamental Theorem of Calculus to simplify the left side

By the Fundamental Theorem of Calculus, the derivative of the integral is equal to the original function: \[ 2^{\frac{1}{2x}} = \frac{d}{dx} \left( k \, 2^{\frac{1}{2x} } + c \right) \]
04

Differentiate the right side of the equation

We will use the chain rule, first differentiating the outer function and then multiplying by the derivative of the inner function. Differentiating the exponential function and applying the chain rule we have: \[2^{\frac{1}{2x}} = k\frac{d}{dx}\left(2^{\frac{1}{2x}}\right) + \frac{d}{dx}\,(c) \] The derivative of a constant is zero, so we only need to differentiate the exponential function: \[2^{\frac{1}{2x}} = k\frac{d}{dx}\left(2^{\frac{1}{2x}}\right)\]
05

Apply the chain rule to differentiate the exponential function

Let \(u = \frac{1}{2x}\), and apply the chain rule: \[\frac{du}{dx} = -\frac{1}{2x^2}\] Now, the differential of 2^u with respect to x will be, using the chain rule: \[\frac{d}{dx}\left(2^{u}\right) = 2^u \cdot \frac{du}{dx} = 2^{\frac{1}{2x}} \cdot -\frac{1}{2x^2}\] So, the equation becomes: \[2^{\frac{1}{2x}} = k 2^{\frac{1}{2x}} \cdot -\frac{1}{2x^2}\]
06

Solve for k

To solve for k, we will divide both sides of the equation by \(2^{\frac{1}{2x}}\cdot -\frac{1}{2x^2}\): \[k = \frac{2^{\frac{1}{2x}}}{2^{\frac{1}{2x}}\cdot -\frac{1}{2x^2}}\] Simplify the equation to isolate k: \[k = -\left(\frac{1}{2x^2}\right)^{-1} = -\frac{1}{2\log{2}}\] So the correct answer is: (a) \(-\frac{1}{2\log{2}}\)

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Most popular questions from this chapter

\(\int\left[\left(\sec ^{2} x-2009\right) /\left(\sin ^{2009} x\right)\right] d x=\) \(-c\) (a) \(\left[(\cot x) /\left(\sin ^{2009} x\right)\right]\) (b) \(\left[(-\cot x) /\left(\sin ^{2009} x\right)\right]\) (c) \(\left[(\tan x) /\left(\sin ^{2009} x\right)\right]\) (d) \(\left[(-\tan \mathrm{x}) /\left(\sin ^{2009} \mathrm{x}\right)\right]\)

If \(\int\left[\left(5^{x} d x\right) / \sqrt{\left(25^{x}-1\right)}\right]=k \log \mid 5^{x}+\sqrt{\left(25^{x}-1\right) \mid+c \text { then }}\) \(\mathrm{k}=\) (a) \(\log _{\mathrm{e}}^{(1 / 5)}\) (b) \(\left[1 /\left(\log _{\mathrm{e}} 5\right)\right]\) (c) \(\log _{\mathrm{e}} 25\) (d) \(\log _{\mathrm{e}}^{(1 / 25)}\)

\(\int \sin x \cdot \cos x \cdot \cos 2 x \cdot \cos 4 x \cdot \cos 8 x \cdot \cos 16 x d x=\ldots+c\) (a) \([(\sin 16 \mathrm{x}) /(1024)]\) (b) \(-[(\cos 32 \mathrm{x}) /(1024)]\) (c) \([(\cos 32 \mathrm{x}) /(1096)]\) (d) \(-[(\cos 32 \mathrm{x}) /(1096)]\)

\(\int \mathrm{e}^{2 \mathrm{x}+\log \mathrm{x}} \mathrm{dx}=\) (a) \((1 / 4)(2 \mathrm{x}-1) \mathrm{e}^{2 \mathrm{x}}\) (b) \((1 / 2)(2 \mathrm{x}-1) \mathrm{e}^{2 \mathrm{x}}\) (c) \((1 / 4)(2 \mathrm{x}+1) \mathrm{e}^{2 \mathrm{x}}\) (d) \((1 / 4)(2 \mathrm{x}+1) \mathrm{e}^{2 \mathrm{x}}\)

$$ \begin{aligned} &\int\left[(\mathrm{dx}) /\left\\{(\mathrm{x}-1)^{3 / 2}(\mathrm{x}-2)^{1 / 2}\right\\}\right]=\underline{\mathrm{c}}\\\ &\text { (a) } 2 \sqrt{[}(\mathrm{x}-1) /(\mathrm{x}-2)] \end{aligned} $$
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