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Let function F be defined as f(x)= int...

Let function F be defined as ` f(x)= int_1^x e^t/t dt x > 0 ` then the value of the integral `int_1^1 e^t/(t+a) dt ` where ` a > 0 ` is

A

`e^(a)[F(x)-F(1+a)]`

B

`e^(-a)[F(x+a)-F(a)]`

C

`e^(a)[F(x+a)-F(1+a)]`

D

`e^(-a)[F(x+a)-F(1+a)]`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the given problem, we need to evaluate the integral: \[ I = \int_1^{1} \frac{e^t}{t + a} \, dt \] where \( a > 0 \). ### Step-by-step Solution: 1. **Understanding the Integral**: The integral is defined from 1 to 1, which means the limits of integration are the same. Therefore, any integral evaluated over the same limits will yield 0. \[ I = \int_1^{1} \frac{e^t}{t + a} \, dt = 0 \] 2. **Conclusion**: Since the upper limit and lower limit of the integral are the same, the value of the integral is: \[ I = 0 \] ### Final Answer: The value of the integral \( \int_1^{1} \frac{e^t}{t + a} \, dt \) is \( 0 \).
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