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What is #int x^4 + 9 x^3 -8 x^2 -3 x + 5 dx#?

Answer 1

#1/5x^5+9/4x^4-8/3x^3-3/2x^2+5x+C#

Use the rule:

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

Also, recall that multiplicative constants (like the #9# in #9x^3#) can be brought outside of the integral, but just stay and are multiplied in, and that various antiderivatives can be added.

Also, the antiderivative of a constant, like #5#, is simply #5x#. This is true since #5=5x^0#, to which the original rule can be applied.

The antiderivative of the given function is:

#x^(4+1)/(4+1)+(9x^(3+1))/(3+1)-(8x^(2+1))/(2+1)-(3x^(1+1))/(1+1)+(5x^(0+1))/(0+1)+C#

#=1/5x^5+9/4x^4-8/3x^3-3/2x^2+5x+C#

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Answer 2

William Abdalla

The integral of (x^4 + 9x^3 - 8x^2 - 3x + 5) with respect to (x) is:

[\frac{1}{5}x^5 + \frac{9}{4}x^4 - \frac{8}{

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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.