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How do you evaluate the integral #intx^3+4x^2+5 dx#?

Answer 1

Because this equation only consists of terms added together, you can integrate them separately and add the results, giving us:

#int x^3 + 4x^2 + 5dx = intx^3dx + int4x^2dx + int5dx#

Each of these terms can be integrated using the Power Rule for integration, which is:

#int x^ndx = x^(n+1)/(n+1) + C#

Plugging our 3 terms into this formula, we have:

#int x^3dx = x^(3+1)/(3+1) = x^4/4#

#int 4x^2dx = (4x^(2+1))/(2+1) = (4x^3)/3#

#int 5dx = int 5x^0dx = (5x^(0+1))/(0+1) = (5x^1)/1 = 5x#

Now we arrive at our final answer by adding these together, remembering to add our constant (#C#) on the end:

#int x^3 + 4x^2 + 5dx = x^4/4 + (4x^3)/3 + 5x + C#

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Answer 2

Isaiah Allen

To evaluate the integral ∫(x^3 + 4x^2 + 5) dx, you can use the power rule for integration. Integrating each term separately, you get ∫x^3 dx + ∫4x^2 dx + ∫5 dx. Applying the power rule, you get (1/4)x^4 + (4/3)x^3 + 5x + C, where C is the constant of integration. Therefore, the integral of x^3 + 4x^2 + 5 dx is (1/4)x^4 + (4/3)x^3 + 5x + C.

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Answer from HIX Tutor

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.