area with inverse

Practise area with inverse with an original VectorPaths question, worked explanation and related VCE Methods practice.

3 practice marks

Original practice question

The area of the region bounded by the curve with equation $y = kx^{\frac{1}{3}}$, where $k$ is a positive constant, the $x$-axis and the line with equation $x = 8$ is 24. Find $k$.

Worked explanation

• Set up the integral

  \[A = \int_0^8 kx^{\frac{1}{3}} \, dx\]

• Evaluate the integral

  \[A = k \left[\frac{x^{\frac{4}{3}}}{\frac{4}{3}}\right]_0^8 = k \cdot \frac{3}{4} \left[x^{\frac{4}{3}}\right]_0^8 = \frac{3k}{4} \cdot 8^{\frac{4}{3}} = \frac{3k}{4} \cdot 16 = 12k\]

• Solve for k

  \[12k = 24 \implies k = 2\]

$k = 4$

Original VectorPaths practice; not an official VCAA question or marking rubric.