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If A(t)=1te|×|dx, then limtA(t) is equal to 

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a
2−e−1
b
3−e−1
c
4
d
0

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detailed solution

Correct option is A

A(t)=∫−10 exdx+∫0t e−xdx=1−e−1−e−t−1limt→∞ A(t)=2−e−1


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