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Evaluate int(40)^(50)(d theta)/(theta - ...

Evaluate `int_(40)^(50)(d theta)/(theta - theta_0),` where `theta_0 = 30^@C,` a constant temperature of surroundings.

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Verified by Experts

The correct Answer is:
-0.6931
`int_(50)^(40) (d theta)/(theta - theta_0) = [log_e ( theta - theta_0)]_(50)^(40) = log_e (40 - 30)/(50 -30) = log_e (1)/(2)`
`=2.3026[log 1 -log2] = 2.3026 [ 0.3010] = - 0.6931.`
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