If I - - 168 -x - -32x - 256-x - -32x - 256 dx then I16 2 is

I = ∫ ^16_8 (| √(x 8)) + 2 √(2) | + | √(x 8) 2 √(2)| dx ⇒ I = ∫ ^16_8√(x 8) + 2 √(2) + 2 √(2) √(x 8) dx ⇒ I = 8× 4 √(2) = 32 √(2) ⇒(I16)^2 = (32 √(2 ...

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Byju's Answer

Standard XII

Mathematics

Definite Integral as Limit of Sum

If I = ∫ 168 ...

Question

If $I = 16 \int 8 (\sqrt{x + \sqrt{32 x - 256}} + \sqrt{x - \sqrt{32 x - 256}}) d x$ then ${(\frac{I}{16})}^{2}$ is

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Solution

$I = 16 \int 8 (| \sqrt{x - 8}) + 2 \sqrt{2} | + | \sqrt{x - 8} - 2 \sqrt{2} | d x \Rightarrow I = 16 \int 8 \sqrt{x - 8} + 2 \sqrt{2} + 2 \sqrt{2} - \sqrt{x - 8} d x \Rightarrow I = 8 \times 4 \sqrt{2} = 32 \sqrt{2} \Rightarrow {(\frac{I}{16})}^{2} = {(\frac{32 \sqrt{2}}{16})}^{2} = 8$

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If I=16∫8(√x+√32x−256+√x−√32x−256) dx then (I16)2 is

I=16∫8(|√x−8)+2√2|+|√x−8−2√2| dx⇒I=16∫8√x−8+2√2+2√2−√x−8 dx⇒I=8×4√2=32√2⇒(I16)2=(32√216)2=8

If $I = 16 \int 8 (\sqrt{x + \sqrt{32 x - 256}} + \sqrt{x - \sqrt{32 x - 256}}) d x$ then ${(\frac{I}{16})}^{2}$ is

$I = 16 \int 8 (| \sqrt{x - 8}) + 2 \sqrt{2} | + | \sqrt{x - 8} - 2 \sqrt{2} | d x \Rightarrow I = 16 \int 8 \sqrt{x - 8} + 2 \sqrt{2} + 2 \sqrt{2} - \sqrt{x - 8} d x \Rightarrow I = 8 \times 4 \sqrt{2} = 32 \sqrt{2} \Rightarrow {(\frac{I}{16})}^{2} = {(\frac{32 \sqrt{2}}{16})}^{2} = 8$