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I'm trying to solve and approximate the area between the two graphs. Right now, my functions are stored as

f[x_] := 3 Sin[x]
g[x_] := x - 1

and then I tried to integrate by evaluating

Integrate[Abs[f[x] - g[x]], x]

Instead of getting an answer, I just get the exact same thing I inputted

Integrate[Abs[f[x] - g[x]], x]

How do I fix this?

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  • $\begingroup$ You can format inline code and code blocks by selecting the code and clicking the {} button above the edit window. The edit window help button ? is useful for learning how to format your questions and answers. You may also find this meta Q&A helpful $\endgroup$ – Michael E2 Apr 12 at 2:35
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Use Assumptions:

Integrate[Abs[f[x] - g[x]], x, Assumptions -> x \[Element] Reals]

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Or try RealAbs instead of Abs:

Integrate[RealAbs[f[x] - g[x]], x]

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(They are equivalent antiderivatives.)

To get the area between the graphs, you need also to solve for the points of intersection.

area = Integrate[
 Abs[f[x] - g[x]], {x, Sequence @@ MinMax[x /. Solve[f[x] == g[x], x, Reals]]}]

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The area is approximately:

N[area]
(*  5.57475  *)
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  • $\begingroup$ RealAbs is awesome to know about! :O $\endgroup$ – Kagaratsch Apr 12 at 2:40
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You need to add assumptions, like this

 Integrate[Abs[f[x] - g[x]], x, Assumptions :> Element[x, Reals]]

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Assuming your functions

f[x_] := 3 Sin[x] 
g[x_] := x - 1

are real valued, you can use square root of square to parametrize the absolute value. This then gives:

Integrate[Sqrt[(f[x] - g[x])^2], x]

(((-2 + x) x + 6 Cos[x]) Sqrt[(-1 + x - 3 Sin[x])^2])/(2 (-1 + x - 3 Sin[x]))

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