How do you find the integral of #int (1/(1+x^2) ) dx# from negative infinity to 0?
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graph{arctanx [-20, 20, -3.12, 3.12]}
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To find the integral of ( \int_{-\infty}^{0} \frac{1}{1+x^2} , dx ), you can use the following approach:
- Recognize that the integrand is a standard form of the arctangent function.
- Apply the property of the arctangent function: ( \int \frac{1}{1+x^2} , dx = \arctan(x) + C ).
- Evaluate the definite integral using the limits of integration: ( \arctan(0) - \lim_{a \to -\infty} \arctan(a) ).
- Since ( \arctan(0) = 0 ) and ( \lim_{a \to -\infty} \arctan(a) = -\frac{\pi}{2} ), the result is ( 0 - (-\frac{\pi}{2}) = \frac{\pi}{2} ).
Therefore, the integral of ( \int_{-\infty}^{0} \frac{1}{1+x^2} , dx ) equals ( \frac{\pi}{2} ).
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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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