If I1-12x-x-3-xdx and I2-12-x-3-xdx- then I1-I2 -

The correct option is B 32 I_1=∫_1^2x(√(x)+√(3 x))dx I_1=∫_1^2(3 x)(√(x)+√(3 x)) dx I_1 =3I_2 nbsp; nbsp;I_1 2I_1 nbsp;= ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Factorization Method Form to Remove Indeterminate Form

If I1=∫12x√...

Question

If $I_{1} = \int_{1}^{2} x (\sqrt{x} + \sqrt{3 - x}) d x$ and $I_{2} = \int_{1}^{2} (\sqrt{x} + \sqrt{3 - x}) d x$ , then $\frac{I_{1}}{I_{2}} =$

\frac{1}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{3}{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

Suggest Corrections

Similar questions

I_{1} = \int_{0}^{1} \frac{{tan}^{- 1} x d x}{x}

I_{2} = \int_{0}^{\frac{π}{2}} \frac{x d x}{sin x}

\frac{I_{1}}{I_{2}}

I_{1} = \int_{e}^{e^{2}} \frac{d x}{l o g x}

I_{2} = \int_{1}^{2} \frac{e^{x}}{x} d x,

f (x) = \frac{e^{x}}{1 + e^{x}}

I_{1} = \int_{f (- a)}^{f (a) x g {x (1 - x)}} d x

I_{2} = \int_{f (- a)}^{f (a) g {x (1 - x)}} d x

\frac{I_{2}}{I_{1}}

f (x) = \frac{e^{x}}{1 + e^{x}}

I_{1} = \int_{f (- a)}^{f (a)} x g {x (1 - x)} d x

I_{2} = \int_{f (- a)}^{f (a)} g {x (1 - x)} d x

\frac{I_{2}}{I_{1}}

f (x) = \frac{e^{x}}{1 + e^{x}}

I_{1} = \int_{f (- a)}^{f (a)} x . g {x (1 - x)} d x

I_{2} = \int_{f (- a)}^{f (a)} g {x (1 - x)} d x

I_{2} / I_{1}

Join BYJU'S Learning Program