The value of is equal to -36-x-12 x-36-x-12 x-36 dxis equal to

The correct option is B 6 √(3)∫_0^3 (√( )sqrt(x+3) + 2 √(3)√(x) + √((x+3) 2 √(3)√(x)) )dx ∫_0^3 ((√(x) + √(3)) + (√(3) √(x))) dx = ∫_0^3 2 √(3) dx = 6 √(3) ...

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Byju's Answer

Standard XII

Mathematics

Second Fundamental Theorem of Calculus

The value of ...

Question

The value of is equal to $\int_{3}^{6} (\sqrt{x + \sqrt{12 x - 36}} + \sqrt{x - \sqrt{12 x - 36}}) d x$ is equal to

6 \sqrt{3}

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4 \sqrt{3}

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12 \sqrt{3}

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2 \sqrt{3}

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Solution

The correct option is A $6 \sqrt{3}$
$\int_{0}^{3} (\sqrt{} s q r t (x + 3) + 2 \sqrt{3} \sqrt{x} + \sqrt{(x + 3) - 2 \sqrt{3} \sqrt{x}}) d x \int_{0}^{3} ((\sqrt{x} + \sqrt{3}) + (\sqrt{3} - \sqrt{x})) d x = \int_{0}^{3} 2 \sqrt{3} d x = 6 \sqrt{3}$

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Similar questions

Q. The value of is equal to

\int_{3}^{6} (\sqrt{x + \sqrt{12 x - 36}} + \sqrt{x - \sqrt{12 x - 36}}) d x

is equal to

Q. The value of

\int_{0}^{6} (\sqrt{x + \sqrt{12 x - 36}} + \sqrt{x - \sqrt{12 x - 36}}) d x

is equal to

Q. The integral

\int_{2}^{4} \frac{log x^{2}}{log x^{2} + log (36 - 12 x + x^{2})}

is equal to

Q. What is the value of x*x+12x+36 when x= 64

Q. The integral

\int_{2}^{4} \frac{log x^{2}}{log x^{2} + log (36 - 12 x + x^{2})} d x

is equal to :

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The value of is equal to ∫63(√x+√12x−36+√x−√12x−36)dxis equal to

The correct option is A 6√3∫30(√ sqrt(x+3)+2√3√x+√(x+3)−2√3√x)dx∫30((√x+√3)+(√3−√x))dx=∫302√3 dx=6√3

The value of is equal to $\int_{3}^{6} (\sqrt{x + \sqrt{12 x - 36}} + \sqrt{x - \sqrt{12 x - 36}}) d x$ is equal to

The correct option is A $6 \sqrt{3}$
$\int_{0}^{3} (\sqrt{} s q r t (x + 3) + 2 \sqrt{3} \sqrt{x} + \sqrt{(x + 3) - 2 \sqrt{3} \sqrt{x}}) d x \int_{0}^{3} ((\sqrt{x} + \sqrt{3}) + (\sqrt{3} - \sqrt{x})) d x = \int_{0}^{3} 2 \sqrt{3} d x = 6 \sqrt{3}$