How do you evaluate the indefinite integral #int (x^2-x+5)dx#?
Remember that an indefinite integral is an antiderivative, and since the derivative of sums is the sum of derivatives then
it is the same for integrals
In this case
Then put them together
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To evaluate the indefinite integral ∫(x^2 - x + 5) dx, you integrate each term separately with respect to x.
∫x^2 dx = (1/3)x^3 + C
∫-x dx = -(1/2)x^2 + C
∫5 dx = 5x + C
Putting it all together:
∫(x^2 - x + 5) dx = (1/3)x^3 - (1/2)x^2 + 5x + C
Where C is the constant of integration.
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To evaluate the indefinite integral ( \int (x^2 - x + 5) , dx ), integrate each term of the polynomial separately. The integral of ( x^2 ) is ( \frac{1}{3}x^3 ), the integral of ( -x ) is ( -\frac{1}{2}x^2 ), and the integral of ( 5 ) is ( 5x ). Combine these results to find the indefinite integral of the given expression. Therefore, ( \int (x^2 - x + 5) , dx = \frac{1}{3}x^3 - \frac{1}{2}x^2 + 5x + C ), where ( C ) is the constant of integration.
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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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