How do you evaluate the integral of #int e^(ln(x^(2)+1))dx# from -2 to -1?
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To evaluate the integral of ( \int e^{\ln(x^2 + 1)} , dx ) from -2 to -1, we can first simplify the expression inside the exponential function using properties of logarithms:
( \ln(x^2 + 1) = \ln((-x)^2 + 1) = \ln(x^2 + 1) )
Now, we can rewrite the integral as:
( \int e^{\ln(x^2 + 1)} , dx = \int (x^2 + 1) , dx )
Integrating ( (x^2 + 1) ) with respect to ( x ) gives:
( \int (x^2 + 1) , dx = \frac{x^3}{3} + x + C )
To evaluate the definite integral from -2 to -1, we substitute the upper limit (-1) and the lower limit (-2) into the antiderivative and subtract:
( \left[\frac{x^3}{3} + x\right]_{-2}^{-1} )
Substituting the upper limit:
( \left[\frac{(-1)^3}{3} - 1\right] )
Substituting the lower limit:
( -\frac{(-2)^3}{3} + (-2) )
Simplify both expressions:
( \left[-\frac{1}{3} - 1\right] = -\frac{4}{3} - 1 )
( -\frac{-8}{3} - 2 = \frac{8}{3} - 2 )
Now, subtracting:
( \frac{8}{3} - 2 = \frac{8}{3} - \frac{6}{3} = \frac{2}{3} )
Therefore, the value of the integral ( \int e^{\ln(x^2 + 1)} , dx ) from -2 to -1 is ( \frac{2}{3} ).
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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