Integration: Question 6

Worked solution

Step 1: Rewrite the integrand as a power of x

$3x^{-1/2}$

is undefined at $x=0$ (division by zero), but as $x\to0^+$ the function grows without bound slowly enough that the area under it up to $x=0$ is still finite. This is exactly the kind of “simple improper case” referred to in the syllabus.

Step 2: Integrate using the power rule

$\int 3x^{-1/2}\,dx = 3 \times \frac{x^{1/2}}{1/2} = 3 \times 2x^{1/2} = 6x^{1/2} = 6\sqrt{x}$

Step 3: Evaluate at the limits

At $x=4$: $\ 6\sqrt{4} = 6(2) = 12$

At $x=0$: $\ 6\sqrt{0} = 0$

$\int_0^4 3x^{-1/2}\,dx = 12 - 0 = 12$

Step 4: Recompute independently by differentiating the antiderivative

$\frac{d}{dx}\left(6x^{1/2}\right) = 6 \times \frac12 x^{-1/2} = 3x^{-1/2}$

This matches the original integrand exactly, confirming $6\sqrt{x}$ is the correct antiderivative and the value $12$ is correct.

Why the other options are wrong

• B ($-12$): comes from dividing by the old exponent $-\frac12$ instead of the new exponent $\frac12$, flipping the sign.
• C ($3$): comes from dividing the coefficient $3$ by $2$ instead of by $\frac12$, giving $1.5x^{1/2}$, which evaluates to $3$ at $x=4$.
• D ($16$): comes from misreading the power as $x^{1/2}$ instead of $x^{-1/2}$, giving antiderivative $2x^{3/2}$, which evaluates to $16$ at $x=4$.

Final answer

$\boxed{\int_0^4 3x^{-1/2}\,dx = 12} \quad \text{(Option A)}$