Solved: The graphs of $y=6-x$ and $y=1.5 x^{2}-2.5 x+3$ intersect at $(2,4)$ and... [IB Mathematics Applications & Interpretation (Math AI)]

Question

SLPaper 1

The graphs of $y=6-x$ and $y=1.5 x^{2}-2.5 x+3$ intersect at $(2,4)$ and $(-1,7)$ , as shown in the following diagrams. In diagram 1, the region enclosed by the lines $y=6-x, x=-1, x=2$ and the $x$ -axis has been shaded. In diagram 2, the region enclosed by the curve $y=1.5 x^{2}-2.5 x+3$ , and the lines $x=-1$ , $x=2$ and the $x$ -axis has been shaded.

Calculate the area of the shaded region in diagram 1.

[3]

Verified

Solution

We can split into right angle triangle and rectangle M1
Triangle Area can be calculated by

$\frac{b\times h}{2}=\frac{(2-(-1))\times (7-4)}{2}=\frac{9}{2}=4.5$ A1

Area of the rectangle can be calculated by $l\times w$

$4\times (2-(-1)=12$

So total area is $16.5$ A1

(other methods like integral or trapzoid earn full marks too)

Write down an integral for the area of the shaded region in diagram 2.

[1]

Verified

Solution

Area below $g(x)=1.5x^2-2.5x+3$ so

A=\int_{-1}^2(1.5x^2-2.5x+3)\,dx

Calculate the area of this region.

[3]

Verified

Solution

Integral from part (b) can be integrated too

\begin{align*} \left[\frac{1.5x^3}{3} - \frac{2.5x^2}{2} + 3x\right]_{-1}^2 &=\left[\frac{1}{2}x^3-\frac{5}{4}x^2+3x\right]_{-1}^2 \\ &=\frac{1}{2}2^3-\frac{5}{4}2^2+3(2)\\ &-\left(\frac{1}{2}(-1)^3-\frac{5}{4}(-1)^2+3(-1)\right) \\ &=4-5+6+\frac{1}{2}+\frac{5}{4}+3 \\ &=9.75 \end{align*}

M1 A1 A1

Hence, determine the area enclosed between $y=6-x$ and $y=1.5 x^{2}-2.5 x+3$ .

[1]

Verified

Solution

16.5 - 9.75 = 6.75

The graphs of $y=6-x$ and $y=1.5 x^{2}-2.5 x+3$ intersect at $(2,4)$ and $(-1,7)$, as shown in the following diagrams. In diagram 1, the region enclosed by the lines $y=6-x, x=-1, x=2$ and the $x$-axis has been shaded.

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