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AP EAMCET Engineering 2018 Apr 22 Shift 2 Paper

Section: Mathematics

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Question : 77 of 160

Marks: +1, -0

\int \left| x^2-3x+2 \right| dx =

$\frac{3}{2}$
$\frac{1}{6}$
$\frac{11}{6}$
$\frac{11}{2}$

Solution:

From given expression,

I=\int\limits_{0}^{3} \left| x^2-3x+2 \right| dx

As,

x^2-3x+1=(x-2)(x-1)

x^2-3x+1<0, x \in (1,2)

And

x^2-3x+2 \ge 0, x \in R - (1,2)

So,

I=\left[ \int\limits_{0}^{1} (x^2-3x+2) dx - \int\limits_{1}^{2} (x^2-3x+2) dx \right]

\left[ + \int\limits_{2}^{3} (x^2-3x+2) dx \right]

=

\left[ \left[ \frac{x^3}{3} - \frac{3x^2}{2} + 2x \right]_{0}^{1} - \left[ \frac{x^3}{3} - \frac{3x^2}{2} + 2x \right]_{1}^{2} \right]

+ \left[ \frac{x^3}{3} - \frac{3x^2}{2} + 2x \right]_{2}^{3}

=

\left[ \left( \frac{1}{3} - \frac{3}{2} + 2 \right) - \left( \frac{8}{3} - \frac{12}{2} + 4 \right) \right]

- \left[ \left( \frac{1}{3} - \frac{3}{2} + 2 \right) \right] + \left[ \left( \frac{27}{3} - \frac{27}{2} + 6 \right) \right]

- \left( \frac{8}{3} - \frac{12}{2} + 4 \right)

= \frac{11}{6}

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