Question

f∫|x|dx limits are upper 2 and lower -2​

Answer 1

$\large\underline\blue{\bold{Given \:  Question :-  }}$

$\tt \:  ⟼\int_{ - 2}^2 \: |x| dx$

$\large\underline\purple{\bold{Solution :- }}$

☆ Let we first define the modulus function.

$\begin{gathered}\begin{gathered}\bf |x| = \begin{cases} &\sf{ - x \: \: \: when \: - 2 \leqslant x \leqslant 0} \\ &\sf{x \: \: \: when \: 0 \leqslant x \leqslant 2} \end{cases}\end{gathered}\end{gathered}$

$\tt \:  ⟼ \: \tt \:  Consider \: \int_{ - 2}^2 \: |x| dx$

$\tt \:  = \: \int_{ - 2}^0 \: |x| dx + \int_{ 0}^2 \: |x| dx$

$\tt \:  = \int_{ - 2}^0 \: - x \: dx \: + \: \int_{ 0}^2 \: x dx$

$\tt \:  = - [\dfrac{ { \: \: x}^{2} }{ 2} ]_{ - 2}^0 \: + \: [\dfrac{ { \: x}^{2} }{2} ]_0^2$

$\tt \:  = - \dfrac{1}{2} [0 - 4 ] + \dfrac{1}{2} [4 - 0]$

$\tt \:  = \: 2 + 2$

$\tt \:  = 4$

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