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## Question of the Week

The diagram shows the curve $y=x^2$ and the straight line $y=0.5x+2$. Find the shaded area shown in purple to 3 decimal places. Solution:

The area shown in purple can be found by subtracting the area beneath the quadratic from the trapezium beneath the straight line. To find the area beneath the quadratic we use definite integrals but we must first identify the limits. Clearly, the lower limit is 0. The upper limit, however, needs to be found by solving $x^2=0.5x+2$, that is, finding the x-coordinate of the intersection of the two graphs: $x^2-0.5x-2=0\hspace{15pt}\Longrightarrow\hspace{15pt}x=-1.186..., 1.686....$

See solving quadratics. Taking the positive x-coordinate, we have the area beneath the quadratic as the following definite integral: $\int_0^{1.686} x^2\,dx=\left[\frac{1}{3}x^3\right]_0^{1.686}=1.598...$

to 3 decimal places. The area of the trapezium (half the sum of the parallel sides multiplied by the distance between them) is given by $0.5\times(2+0.5\times 1.686+2)\times1.686=4.083...$

to 3 decimal places. Hence, the area of the region shaded in green is 4.083-1.598=2.485 square units to 3 decimal places.

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