I=∫_0^4 |x^2 4x+3|dx =2∫_0^2 |x^2 4x+3| dx Now, x^2 4x+3= x ^ 2 4x+3 0≤ x≤ 1 ( x ^ 2 4x+3 ) 1≤ x≤ 2 Hence, 12I=∫_0^2 |x^2 ...

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Integration of Irrational Algebraic Fractions - 2

∫04 |x2 - 4x ...

Question

$4 \int 0 | x^{2} - 4 x + 3 | d x$

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Solution

$I = \int_{0}^{4} | x^{2} - 4 x + 3 | d x$

$= 2 \int_{0}^{2} | x^{2} - 4 x + 3 | d x$

Now, $x^{2} - 4 x + 3 = {\begin{matrix} x^{2} - 4 x + 30 \leq x \leq 1 - (x^{2} - 4 x + 3) 1 \leq x \leq 2 \end{matrix}$

Hence, $\frac{1}{2} I = \int_{0}^{2} | x^{2} - 4 x + 3 | d x$
$= \int_{0}^{1} | x^{2} - 4 x + 3 | d x + \int_{1}^{2} | x^{2} - 4 x + 3 | d x$

$= \int_{0}^{1} (x^{2} - 4 x + 3) d x + \int_{1}^{2} - (x^{2} - 4 x + 3) d x$
$= {[\frac{1}{3} x^{3} - 2 x^{2} + 3 x]}_{0}^{3} - {[\frac{1}{3} x^{3} - 2 x^{2} + 3 x]}_{1}^{3}$

$= (\frac{1}{3} - 2 + 3) - {(\frac{8}{3} - 8 + 6) - (\frac{1}{3} - 2 + 3)}$

$∴ I = 4$

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