Plot z as a function of "a"?

[fernandsneto (Jose richard)]

How can I plot z as a function of “a”?

z = Table[NIntegrate[Sin[a + b x], {x, -10, 10}], {b, 0, 5, 0.5}, {a, 0, 5,  0.5}];
ListPlot[z, Joined -> True, PlotRange -> All]

[trounev (Alexander Trounev)]

u = Integrate[Sin[a + b*x], {x, -10, 10}];
{Plot3D[u, {a, 0., 5}, {b, 0.0001, 5}, PlotRange → All, Mesh → None,
ColorFunction → Hue, AxesLabel → {“a”, “b”, “”}],
Plot[u /. b → 1, {a, 0, 5}, AxesLabel → Automatic]}

[fernandsneto (Jose richard)]

Hey. Thank you a lot. Why did you replace NIntegrate with Integratre?

[trounev (Alexander Trounev)]

In[1]:= u = Integrate[Sin[a + b*x], {x, -10, 10}]
Out[1]= (2 Sin[a] Sin[10 b])/b

[fernandsneto (Jose richard)]

Yes, but this function is just a simple example. My original function is more complicated, so I need NIntegrate.

[trounev (Alexander Trounev)]

OK! Let me take a look at the general cases on this particular example:

u = Flatten[
  Table[{a, b, NIntegrate[Sin[x^a - x^b], {x, 1, 2}]}, {a, 1, 
    2, .1}, {b, 3, 4, .1}], 1];
ListPlot3D[u, AxesLabel -> {"a", "b", ""}, Mesh -> None, 
 ColorFunction -> Hue]

Fu = Interpolation[u]
{Plot[Fu[a, 3.5], {a, 1, 2}, AxesLabel -> Automatic], 
 Plot3D[Fu[a, b], {a, 1, 2}, {b, 3, 4}, Mesh -> None, 
  ColorFunction -> Hue]}

[fernandsneto (Jose richard)]

Thank you again. This is exactly equal to

F[a_?NumericQ, b_?NumericQ] := NIntegrate[Sin[x^a - x^b], {x, 1, 2}]
Plot[F[y, 3.5], {y, 1, 2}]

Why that?