JEE Main 2022 · 26 June · Morning Shift · Mathematics · Definite Integration
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f\left(x\right) =max{∣x+1∣,∣x+2∣,∣x+3∣,∣x+4∣,∣x+5∣}max \left\{\left|x + 1\right|, \left|x + 2\right|, \left|x + 3\right|, \left|x + 4\right|, \left|x + 5\right|\right\}max{∣x+1∣,∣x+2∣,∣x+3∣,∣x+4∣,∣x+5∣}
∫−60f(x)dx=∫−6−3∣x+1∣dx+∫−30∣x+5∣dx{∫}_{- 6}^{0}f\left(x\right)dx = {∫}_{- 6}^{- 3}\left|x + 1\right|dx + {∫}_{- 3}^{0}\left|x + 5\right|dx∫−60f(x)dx=∫−6−3∣x+1∣dx+∫−30∣x+5∣dx
=−∫−6−3(x+1)dx+∫−30(x+5)dx= - {∫}_{- 6}^{- 3}\left(x + 1\right)dx + {∫}_{- 3}^{0}\left(x + 5\right)dx=−∫−6−3(x+1)dx+∫−30(x+5)dx
=−[x22+x]−6−3+[x22+5x]−30= - {\left[\dfrac{{x}^{2}}{2} + x\right]}_{- 6}^{- 3} + {\left[\dfrac{{x}^{2}}{2} + 5x\right]}_{- 3}^{0}=−[2x2+x]−6−3+[2x2+5x]−30
=−[(92−3)−(18−6)]+[0−(92−15)]= - \left[\left(\dfrac{9}{2} - 3\right) - \left(18 - 6\right)\right] + \left[0 - \left(\dfrac{9}{2} - 15\right)\right]=−[(29−3)−(18−6)]+[0−(29−15)]
=−[32−12⌉+212=212+212=21= - \left[\dfrac{3}{2} - 12\right⌉ + \dfrac{21}{2} = \dfrac{21}{2} + \dfrac{21}{2} = 21=−[23−12⌉+221=221+221=21
