The value of -4-5e x-5 2dx-3-1-32-3e9 x-2-3 2dx is

The correct option is B none of theseLet I=∫ _ 4 ^ 5 e^( x+5 ) ^ 2 dx+3∫ _ 1/3 ^ 2/3 e^ 9( x 2/3 ) ^ 2 dx= I _ 1 + I _ 2 Where I _ 1 =∫ _ 4 ^ 5 ...

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

Byju's Answer

Standard XII

Mathematics

Sum of Product of Binomial Coefficients

The value of ...

Question

The value of $\int_{- 4}^{- 5} e^{{(x + 5)}^{2}} d x + 3 \int_{1 / 3}^{2 / 3} e^{9 {(x - 2 / 3)}^{2}} d x$ is

2 / 5

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

1 / 5

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

1 / 2

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

none of these

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

Open in App

Solution

The correct option is B none of these
Let $I = \int_{- 4}^{- 5} e^{{(x + 5)}^{2}} d x + 3 \int_{1 / 3}^{2 / 3} e^{9 {(x - 2 / 3)}^{2}} d x = I_{1} + I_{2}$
Where $I_{1} = \int_{- 4}^{- 5} e^{{(x + 5)}^{2}} d x$
Put $x + 5 = t$
$I_{1} = - \int_{0}^{1} e^{t^{2}} d t$
And $I_{2} = 3 \int_{1 / 3}^{2 / 3} e^{9 {(x - 2 / 3)}^{2}} d x$
Put $- 3 x + 2 = u$
$I_{2} = \int_{0}^{1} e^{u^{2}} d u$
$∴ I = 0$

Suggest Corrections

Similar questions

\int_{- 4}^{- 5} e^{(x + 5)^{2}} d x + 3 \int_{\frac{1}{3}}^{\frac{2}{3}} e^{9 (x - \frac{2}{3})^{2}} d x =

\int_{- 4}^{- 5} e^{{(x + 5)}^{2}} d x + \int_{1 / 3}^{2 / 3} e^{{9 (x - \frac{2}{3})}^{2}} d x

is equal to

Q. 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?

If $a = \frac{\sqrt{5} + 1}{\sqrt{5} - 1} a n d b = \frac{\sqrt{5} - 1}{\sqrt{5} + 1}, t h e v a l u e o f (\frac{a^{2} + a b + b^{2}}{a^{2} - a b + b^{2}})$ is :

Q. The value of

\frac{1}{(216)^{- 2 / 3}} + \frac{1}{(256)^{- 3 / 4}} + \frac{1}{(32)^{- 1 / 5}}

is:

Join BYJU'S Learning Program

The value of ∫−5−4e(x+5)2dx+3∫2/31/3e9(x−2/3)2dx is

The correct option is B none of theseLet I=∫−5−4e(x+5)2dx+3∫2/31/3e9(x−2/3)2dx=I1+I2Where I1=∫−5−4e(x+5)2dxPut x+5=tI1=−∫10et2dtAnd I2=3∫2/31/3e9(x−2/3)2dxPut −3x+2=uI2=∫10eu2du∴I=0

The value of $\int_{- 4}^{- 5} e^{{(x + 5)}^{2}} d x + 3 \int_{1 / 3}^{2 / 3} e^{9 {(x - 2 / 3)}^{2}} d x$ is