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in Definite Integrals by (55 points)
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Find the area bounded by the curve y = x² - 4x + 2 and the line y = x-2.

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by (710 points)

Given curve is

y = x2 – 4x + 2

⇒ y = x2 – 4x + 4 – 2

⇒ y = (x – 2)2 – 2

⇒ (x – 2)2 = y + 2

Hence, given curve is a parabola whose vertex is (2, –2).

For intersection point of given curve and line y = x – 2.

we have (x – 2) = x2 – 4x + 2

⇒ x2 – 5x + 4 = 0

⇒ (x – 4)(x – 1) = 0

⇒ x = 1, 4.

Hence, intersection point are (1, –1) & (4, 2).

Intersection point of parabola and y-axis is (0, 2).

For intersection point of parabola and x-axis is,

we have x2 – 4x + 2 = 0

⇒ x \(=\frac{4\pm \sqrt{16-8}}{2}\)

⇒ x \(=\frac{4\pm2\sqrt 2}{2}\)

⇒ x = 2 \(\pm\) \(\sqrt 2\).

Hence, intersection points are (2 – \(\sqrt 2\), 0) & (2 + \(\sqrt 2\), 0).

Required area \(=\displaystyle\int^4_1\)\(\left[(\mathrm x-2)-(\mathrm x-2)^2-2\right]d\mathrm x\)

\(=\displaystyle\int^4_1\)\((\mathrm x-(\mathrm x-2)^2)d\mathrm x\)

\(=\left(\frac{\mathrm x^2}{2}-\frac{1}{3}(\mathrm x-2)^3\right)^4_1\)

\(=\frac{1}{2}(16-1)-\frac{1}{3}(8+1)\)

\(=\frac{15}{2}-3\)

\(=\frac{15-6}{2}\)

\(=\frac{9}{2}\)

Hence, the area of bounded region by given parabola and line is \(\frac{9}{2}\) square unit.

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