Porisma

Worked solution

Split the area at both roots

Find the area of the region bounded by y=x2−6x+8y=x^2-6x+8, the xx-axis and the lines x=1x=1 and x=5x=5.

CJC · Lecture · Example 7

Sketch and split

Factor x2−6x+8=(x−2)(x−4)x^2-6x+8=(x-2)(x-4). The upward-opening parabola crosses the axis at 22 and 44, with vertex (3,−1)(3,-1). Shade the three pieces between x=1x=1 and x=5x=5: above the axis on [1,2][1,2] and [4,5][4,5], below it on [2,4][2,4].

How did this part go?

Form the area integral

The required area is A=∫12(x2−6x+8) dx−∫24(x2−6x+8) dx+∫45(x2−6x+8) dxA=\int_1^2(x^2-6x+8)\,dx-\int_2^4(x^2-6x+8)\,dx+\int_4^5(x^2-6x+8)\,dx. The Teacher calculator screen enters the same three integrals of Y1Y_1, with signs +,−,++,-,+, and displays 44.

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Evaluate and check the sign

An antiderivative is F(x)=x3/3−3x2+8xF(x)=x^3/3-3x^2+8x. The three signed integrals are 4/3,−4/3,4/34/3,-4/3,4/3, giving A=4/3−(−4/3)+4/3=4A=4/3-(-4/3)+4/3=4 square units. Directly taking ∫15f(x) dx=4/3\int_1^5f(x)\,dx=4/3 would allow cancellation and would not give the area.

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Compare your sketch

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Upward parabola with roots 2 and 4 and vertex (3,−1). Between x=1 and x=5, the two positive pieces and the middle negative piece all contribute positive geometric area.
Upward parabola with roots 2 and 4 and vertex (3,−1). Between x=1 and x=5, the two positive pieces and the middle negative piece all contribute positive geometric area.
Upward parabola with roots 2 and 4 and vertex (3,−1). Between x=1 and x=5, the two positive pieces and the middle negative piece all contribute positive geometric area.

Common mistakes

Do not integrate across both roots without changing the sign of the below-axis contribution.

Read the hint