Let f ( x ) = { − 2 , x − 2 , − 2 ≤ x ≤ 0 0 < x ≤ 2 and h ( x ) = f ( ∣ x ∣ ) + ∣ f ( x ) ∣ . Then ∫ − 2 2 h ( x ) d x is equal to :

h ( x ) = { x − 2 + 2 − x = 0 , − x − 2 + 2 = − x 0 ≤ x ≤ 2 − 2 ≤ x < 0 ⇒ ∫ 0 2 h ( x ) dx = 0 and ∫ − 2 0 h ( x ) dx = 2

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MathematicsDefinite IntegrationMiscellaneous/MixedMedium2 minPYQ_2024

MathematicsMediumsingle choice

Let $f (x) = {- 2, x - 2, - 2 \leq x \leq 0 0 < x \leq 2$ and $h (x) = f (∣ x ∣) + ∣ f (x) ∣$ . Then $\int_{- 2}^{2} h (x) d x$ is equal to :

Question diagram: Let f ( x ) = { − 2 , x − 2 , − 2 ≤ x ≤ 0 0 < x ≤ 2 and

Options:

Answer:

Solution:

$h (x) = {x - 2 + 2 - x = 0, - x - 2 + 2 = - x 0 \leq x \leq 2 - 2 \leq x < 0$
$\Rightarrow \int_{0}^{2} h (x) dx = 0$ and $\int_{- 2}^{0} h (x) dx = 2$

Stream:JEESubject:MathematicsTopic:Definite IntegrationSubtopic:Miscellaneous/Mixed

⏱ 2mℹ️ Source: PYQ_2024

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