The correct option is B 4 ∫_0^4|2 x|dx ∫_0^2(2 x)dx+∫_2^4 (2 x)dx ∫_0^2( x)dx+∫_2^4x #160; #160; dx = x^22∫_0^2+x^22∫_2^4 = 2 ...

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Substitution Method to Remove Indeterminate Form

∫04 |2-x| d...

Question

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

0

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4

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\frac{1}{2}

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\frac{3}{2}

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Solution

The correct option is B $4$
$\int_{0}^{4} | 2 - x | d x$
$\int_{0}^{2} (2 - x) d x + \int_{2}^{4} - (2 - x) d x$
$\int_{0}^{2} (- x) d x + \int_{2}^{4} x d x$
$= \frac{- x^{2}}{2} \int_{0}^{2} + \frac{x^{2}}{2} \int_{2}^{4}$
$= - 2 + 8 - 2 = 4$

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Similar questions

\int_{0}^{4} (| x - l | + 3 | x - 2 | + 4 | x - 3 |) d x =

\int_{0}^{4} (| x | + | x - 2 | + | x - 4 |)

I_{1} = \int_{0}^{1} 2^{x^{2}} d x, I_{2} = \int_{0}^{1} 2^{x^{3}} d x, I_{3} = \int_{1}^{2} 2^{x^{2}} d x

I_{4} = \int_{1}^{2} 2^{x^{3}} d x

I_{1} = \int_{0}^{1} 2^{x^{3}} d x, I_{2} = \int_{0}^{1} 2^{x^{2}} d x, I_{3} = \int_{1}^{2} 2^{x^{2}} d x

I_{4} = \int_{1}^{2} 2^{x^{3}} d x

I f I_{1} = \int_{0}^{1} 2^{x^{2}} d x, I_{2} = \int_{0}^{1} 2^{x^{3}} d x, I_{3} = \int_{1}^{2} 2^{x^{2}} d x a n d I_{4} = \int_{1}^{2} 2 x^{3} d x t h e n

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