Can we find the area inside a straight line, a parabola and the x-axis?

Can we find the area inside a straight line, a parabola and the x-axis?

Question

The figure shows the curve $y = x^2$ and the straight line $2x + y = 15$. Find

1. the co-ordinates of $P$ and $Q$,

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1. the area of the shaded region.

Suggestion

The figure shows the curve $y = x^2$ and the straight line $2x + y = 15$. Find […] the area of the shaded region.

Is there a way of viewing the region differently?

A further suggestion

Could we, perhaps, view it as the sum of two simpler regions?

Solution

The figure given in the question is the following.

The point $P$ is the point in the first quadrant at which the curve $y = x^2$ and line $y = 15 - 2x$ cross. That is, it is the point whose $x$-coordinate is positive and such that $x^2 = 15 - 2x.$ This equation can be rearranged to become $x^2 + 2x - 15 = 0$ which, by inspection, factorises: $(x + 5)(x - 3) = 0.$ Thus, the $x$-coordinate of $P$ is $3$, and so the coordinates of $P$ are $(3,9)$ since, in particular, $P$ lies on the curve $y = x^2$.

The point $Q$ is the $x$-intercept of the line, so the $x$-coordinate of $Q$ is the value of $x$ such that $15 - 2x = 0.$ Thus, $x = \tfrac{15}{2}$, and so the coordinates of $Q$ are $\left( \tfrac{15}{2}, 0 \right)$.

The key to this question is to break the region into two pieces, as the following diagram shows.

The shaded area is therefore the sum of the area of region $A$ and the area of region $B$, which are separated by the dotted line segment in the image. This dotted line segment is part of the line $x = 3$, since the $x$-coordinate of the point $P$ is $3$.

The area of region $A$ can be calculated by integration: $\text{area of }A\text{} = \int_0^3 x^2 \:dx = \left[ \frac{x^3}{3} \right]_0^3 = \frac{3^3}{3} - 0 = 9.$

On the other hand, region $B$ is a (right-angled) triangle whose base has length $\frac{15}{2} - 3 = \frac{9}{2}$ (which corresponds to the $x$-coordinate of $P$ being subtracted from the $x$-coordinate of $Q$) and whose height has length $9$, which corresponds to the $y$-coordinate of $P$. Thus, $\text{area of }B\text{} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{9}{2} \times 9 = \frac{81}{4}.$

Therefore, $\text{area of shaded region} = \text{area of }A\text{} + \text{area of }B\text{} = 9 + \frac{81}{4} = \frac{117}{4}.$