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I have this definite integral $$\int_{-1}^14x^3-x\,dx=0$$

I had that function rendered and found out it should be calculated in four intervals:

$[-1,-0.5], [-0.5,0], [0,0.5], [0.5,1].$

Is there any other (shorter and correct) method how to calculate that?

thx :)

Andrew Chin
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naruto25
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  • an antiderivative is $x^4-\frac12 x^2$; that evaluated from $-1 $ to $1$ is $0$ – J. W. Tanner Dec 23 '19 at 01:11
  • thx, but do I have to calculate every subinterval separately in order to get correct result or is there any "shorter method"? Because I cannot just take only interval <-1;1> and calculate definite integral with it. – naruto25 Dec 23 '19 at 01:12
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    why can't you take only $[-1,1]$? Also, note that the integral of an odd function over an interval symmetric about $0$ is $0$ – J. W. Tanner Dec 23 '19 at 01:13
  • Because there are parts under axis X and above axis X. I am not sure, if I can calculate it just straight. – naruto25 Dec 23 '19 at 01:15
  • you can calculate it over the entire interval; it is well-defined over the entire interval – J. W. Tanner Dec 23 '19 at 01:20

2 Answers2

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$$\int_{-1}^1 (4x^3-x)dx=\left(x^4-\frac12x^2\right)|_{-1}^{1}=0$$

If you prefer, $$\int_{-1}^1 (4x^3-x)dx=\int_{-1}^0 (4x^3-x)dx+\int_0^1(4x^3-x)dx$$

$$=\left(x^4-\frac12x^2\right)|_{-1}^{0}+\left(x^4-\frac12x^2\right)|_0^1=-\frac12+\frac12=0,$$

or simply note that the integral of an odd function over an interval symmetric about $0$ is $0 $.

J. W. Tanner
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  • Thx. Ok, so if I have to calculate area, I have to split that interval and calculate absolute value of those intervals, but if I calculate only definite interval, I can leave it as it is? – naruto25 Dec 23 '19 at 01:22
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    The definite integral is the signed area – J. W. Tanner Dec 23 '19 at 01:31
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You should distinguish two questions:

1) Calculate the definite integral: $$\int_{-1}^1 4x^3-x\, dx=\left(x^4-\frac{x^2}{2}\right)|_{-1}^1=\frac12-\frac12=0.$$ 2) Calculate the area between $y=4x^3-x$ and the $x$-axis: $$A=\left|\int_{-1}^{-0.5} 4x^3-x\, dx\right|+\left|\int_{-0.5}^{0} 4x^3-x\, dx\right|+\left|\int_{0}^{0.5} 4x^3-x\, dx\right|+\left|\int_{0.5}^{1} 4x^3-x\, dx\right|=\\\ \left|\left(x^4-\frac{x^2}{2}\right)|_{-1}^{-0.5}\right|+\left|\left(x^4-\frac{x^2}{2}\right)|_{-0.5}^{0}\right|+\left|\left(x^4-\frac{x^2}{2}\right)|_{0}^{0.5}\right|+\left|\left(x^4-\frac{x^2}{2}\right)|_{0.5}^{1}\right|=\\ |-0.0625-0.5|+|0-(-0.0625)|+|-0.0625-0|+|0.5-(-0.0625)|= 1.25.$$

farruhota
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