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Answer any one of the following.
The equation y = 4 - x² represents a parabola. Make a rough sketch of the graph of the given…
The equation $y = 4 - x^2$ represents a parabola.
(a)[1]
Make a rough sketch of the graph of the given function.
Draw: graph of $y = 4 - x^2$
(b)[2]
Determine the area enclosed between the curve, the $x$-axis, the lines $x = 0$ and $x = 2$.
(c)[1]
Hence, find the area bounded by the parabola and the $x$-axis.
Answer
Answer (a)
The graph of $y=4-x^2$ is a downward parabola with vertex $(0,4)$, cutting the $x$-axis at $(-2,0)$ and $(2,0)$ and the $y$-axis at $(0,4)$; it is symmetric about the $y$-axis.
Downward parabola, vertex (0,4), x-intercepts (-2,0) and (2,0), symmetric about the y-axis
Answer (b)
Integral: $\int_{0}^{2} \left(4-x^2\right) dx$
Area $=\int_0^2(4-x^2)\,dx=\left[4x-\frac{x^3}{3}\right]_0^2=8-\frac83=\frac{16}{3}$ sq. units.
Final answer: $\frac{16}{3}$ sq. units
Answer (c)
Integral: $\int_{-2}^{2} \left(4-x^2\right) dx$
The parabola is symmetric about the $y$-axis, so the area from $x=-2$ to $x=2$ is twice that from $0$ to $2$.
Area $=2\times\frac{16}{3}=\frac{32}{3}$ sq. units.
Final answer: $\frac{32}{3}$ sq. units