If I1- -4-5e x -5 2dx and I2- 3-1-32-3e 3x -2 2dx then I1-I2 equals-

The correct option is A 0 I_ 2 =3∫ _ 1 / 3 nbsp;^ 2 / 3 nbsp; e^( 3x 2 ) ^ 2 dx x= y 3 / 3 nbsp; nbsp; nbsp;dx= dy / 3 nbs ...

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Byju's Answer

Standard XII

Mathematics

Factorization Method Form to Remove Indeterminate Form

If I1= ∫-4-...

Question

If $I_{1} = \int_{- 4}^{- 5} e^{{(x + 5)}^{2}} d x$ and $I_{2} = 3 \int_{1 / 3}^{2 / 3} e^{{(3 x - 2)}^{2}} d x$

then $I_{1} + I_{2}$ equals?

\frac{1}{3}

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- \frac{1}{3}

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0

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Solution

The correct option is A $0$
$I_{2} = 3 \int_{\frac{1}{3}}^{\frac{2}{3}} e^{{(3 x - 2)}^{2}} d x$
$x = \frac{- y - 3}{3}$
$d x = - \frac{d y}{3}$
$\Rightarrow I_{2} = 3 \int_{\frac{1}{3}}^{\frac{2}{3}} e^{{(3 x - 2)}^{2}} d x = - \int_{- 4}^{- 5} e^{{(y + 5)}^{2}} d x = - I_{1}$
$\Rightarrow I_{1} + I_{2} = 0$

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If I1=∫−5−4e(x+5)2dx and I2=3∫2/31/3e(3x−2)2dx then I1+I2 equals?

The correct option is A 0I2=3∫2313e(3x−2)2dx x=−y−33 dx=−dy3 ⇒I2=3∫2313e(3x−2)2dx=−∫−5−4e(y+5)2dx=−I1 ⇒I1+I2=0

If $I_{1} = \int_{- 4}^{- 5} e^{{(x + 5)}^{2}} d x$ and $I_{2} = 3 \int_{1 / 3}^{2 / 3} e^{{(3 x - 2)}^{2}} d x$

then $I_{1} + I_{2}$ equals?