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Q.

∫−4−5 e(x+5)2dx+3∫1/32/3 e9x−232dx is

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a

0

b

–2

c

1

d

2

answer is A.

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Detailed Solution

I1=∫−4−5 e(x+5)2dx=−∫01 et2dt(t=x+5)I2=∫1/32/3 e9(x−2/3)2dx=∫1/32/3 e(3x−2)2dx (3x−2=t)=13∫−10 et2dt=−13∫10 et2dt=13∫01 et2dtI1+3I2=0

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∫−4−5 e(x+5)2dx+3∫1/32/3 e9x−232dx is