Exercise 7.1 · Q21 (MCQ): Antiderivative of $\sqrt{x} + \frac{1}{\sqrt{x}}$

Basics you need

Solution

1. Domain

   $\sqrt{x}$ needs $x \ge 0$⁠, and $\frac{1}{\sqrt{x}}$ also needs $\sqrt{x} \ne 0$⁠. So $f$ is defined for $x > 0$⁠, and that is where we look for its anti derivative.

   All four options involve powers like $x^{1/2}$ and $x^{3/2}$⁠, which are real for $x \ge 0$⁠; comparing derivatives on $x > 0$ is therefore fair to every option.
   The Note in Section 7.2: always keep the interval of definition in mind.

2. Write the integrand as powers of $x$

   $$\sqrt{x} + \frac{1}{\sqrt{x}} = x^{1/2} + x^{-1/2}.$$
   The power formula only applies to $x^n$⁠. Moving $\sqrt{x}$ from the denominator to the numerator changes the sign of its exponent.
   $\frac{1}{x^n} = x^{-n}$⁠. Basics: powers of x

3. Integrate with the power formula

   $$\begin{aligned} \int x^{1/2}\,dx &= \frac{x^{3/2}}{3/2} = \frac23 x^{3/2}, \\ \int x^{-1/2}\,dx &= \frac{x^{1/2}}{1/2} = 2x^{1/2}. \end{aligned}$$

   So

   $$\int\left(\sqrt{x} + \frac{1}{\sqrt{x}}\right)dx = \frac23 x^{3/2} + 2x^{1/2} + C.$$
   Raise each power by 1 ($\frac12 \to \frac32$⁠, $-\frac12 \to \frac12$⁠) and divide by the new power, that is, multiply by its reciprocal ($\frac23$ and $2$⁠). Neither exponent is $-1$⁠. Property III adds the two results, with one constant $C$⁠.
   Formula (i), Section 7.2; Property III, Section 7.2.1. Lesson: integration as anti-differentiation

4. Check: differentiate the answer back

   $$\begin{aligned} &\frac{d}{dx}\left(\tfrac23 x^{3/2} + 2x^{1/2} + C\right) \\ &\quad = \tfrac23\cdot\tfrac32 x^{1/2} + 2\cdot\tfrac12 x^{-1/2} \\ &\quad = \sqrt{x} + \frac{1}{\sqrt{x}}. \end{aligned}$$
   The derivative is $f(x)$ for every $x > 0$⁠, so this is an anti derivative of $f$⁠. It is option (C) exactly, including the constant $C$⁠.
   Property I, Section 7.2.1. Basics: integration undoes differentiation

5. Examine every option

   Differentiate each option (the $+C$ contributes 0) and compare with $f(x) = x^{1/2} + x^{-1/2}$⁠. The last column is the derivative at $x = 4$⁠, where $f(4) = 2 + \frac12 = 2.5$⁠.

   Option	Derivative	Equals $f$⁠?	At $x = 4$
   (A)	$\frac19 x^{-2/3} + x^{-1/2}$	No: first term wrong	$\approx 0.5441$
   (B)	$\frac49 x^{-1/3} + x$	No: both terms wrong	$\approx 4.2800$
   (C)	$x^{1/2} + x^{-1/2}$	Yes	$2.5$
   (D)	$\frac94 x^{1/2} + \frac14 x^{-1/2}$	No: coefficients wrong	$4.625$

   (A) keeps the correct second term $2x^{1/2}$ but its first term uses a cube root, $x^{1/3}$⁠, which has nothing to do with $\sqrt{x}$⁠. (B) has the powers $x^{2/3}$ and $x^2$⁠, which differentiate to $x^{-1/3}$ and $x$⁠; neither appears in our integrand. (D) has the right powers but multiplies by the new exponent instead of dividing: $\frac32 x^{3/2}$ differentiates to $\frac94 x^{1/2}$⁠, not $x^{1/2}$⁠. Only (C) passes.
   By definition (Section 7.2), $F$ is an anti derivative of $f$ exactly when $F' = f$⁠, so an option is correct only if its derivative is exactly $f$⁠. The $+\,C$ in every option is harmless: functions with the same derivative differ by a constant (Remark, Section 7.2). Basics: integration undoes differentiation

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