0
$\begingroup$

How do I solve this:approximate the largest and the smallest values of the integral

enter image description here

for

enter image description here

$\endgroup$
3
  • 1
    $\begingroup$ The integrand is positive on the interval $1\leq x \leq 3$ so the extremes certainly happen at the endpoints. Though, perhaps you mean to ask this on math.se? $\endgroup$
    – Mark McClure
    Aug 31, 2014 at 1:25
  • 1
    $\begingroup$ Since the integrand is positive in the given range {1, 3} the smallest value is at $1$ and the largest value is at $3$. Approximate values are NIntegrate[t/Sqrt[2 + t^3], {t, -1, #}] & /@ {1, 3} i.e. {-0.0764949, 1.15208}. $\endgroup$
    – Artes
    Aug 31, 2014 at 1:30
  • $\begingroup$ @MarkMcClure I appreciate I should have commented re: ? Math.se and I have edited my incorrect statements in my 'answer'...was just trying to helpful $\endgroup$
    – ubpdqn
    Aug 31, 2014 at 1:41

1 Answer 1

Reset to default
1
$\begingroup$

Please note that some illustration of your efforts to solve this problem with Mathematica is regarded as a minimum requirement for questions. This aids your skills and facilitates assistance.

For motivation:

Plot[t/Sqrt[2 + t^3], {t, -1, 3}, Filling -> Axis]

enter image description here

For the interval of interest the function is positive and close to constant. The integral of the function will be monotone increasing (and close to linear ), hence the minimum and maximum will be the end points of the interval.

I suggest you program using NIntegrate or Integrate.

Just for 'fun', you can use new functions in v10:

ir0 = ImplicitRegion[0 > y > x/Sqrt[2 + x^3] && -1 < x < 0, {x, y}];
irf[u_] := 
 ImplicitRegion[0 < y < x/Sqrt[2 + x^3] && 0 < x < u, {x, y}]
func[v_] := N[RegionMeasure[irf[v]] - RegionMeasure[ir0]]

Confirming the qualitative observation:

Plot[func[v], {v, 1, 3}]

enter image description here

You can now answer your minimum, maximum question.

$\endgroup$

You must log in to answer this question.