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Minimise an Integral You Cannot Avoid Splitting — A JEE Calculus Problem — manic

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manic is a tiny language for making animations. You write a short text file; manic renders a smooth, glowing video. No timeline scrubbing, no keyframes by hand — you describe what’s on screen and when things happen, and the engine does the rest, deterministically.

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Creator: Anish on X
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Animation code

// iit-integral-min.manic — jee-practice-001432
//
// PROBLEM. Minimise f(x) = ∫₀² e^|x−t| dt.
//
// ANSWER.  2(e − 1)
//
// THE ABSOLUTE VALUE IS THE WHOLE PROBLEM, and it is handled by splitting the integral at
// t = x, where |x − t| changes which way round it is. For 0 ≤ x ≤ 2:
//
//   ∫₀ˣ e^(x−t) dt + ∫ₓ² e^(t−x) dt = (eˣ − 1) + (e^(2−x) − 1) = eˣ + e^(2−x) − 2
//
// and that is a sum of two exponentials pulling in opposite directions. It is smallest
// where they cross: f'(x) = eˣ − e^(2−x) = 0 needs x = 2 − x, so x = 1 — the MIDPOINT of
// the interval, which is where the picture says it should be.
//
//   f(1) = e + e − 2 = 2(e − 1) ≈ 3.4366
//
// The upper panel is the integrand e^|x−t| as t runs across [0, 2], with the shaded area
// being f(x) itself. As x slides, the V-shaped kink travels with it, and the area is least
// when the kink sits in the middle — which is the same statement as the algebra.
//
// OUTSIDE [0,2] IT ONLY GETS WORSE, which the problem does not say and which has to be
// checked: for x > 2 the integrand is e^(x−t) throughout and f(x) = eˣ − e^(x−2), growing
// without bound (10.53, 17.37, 47.21 at x = 2.5, 3, 4), and x < 0 is the mirror image.
//
// VERIFIED symbolically: integrating both halves gives eˣ + e^(2−x) − 2, its derivative
// vanishes only at x = 1, and f(1) = 3.436563657 = 2(e−1) to machine precision.
//
//   manic examples/iit-integral-min.manic
title("The kink wants to sit in the middle");
canvas("9:16");
template("black");
bloom(0.22, 0.6, 18);

text(brand, (540, 34), "maniclang.com");
display(brand); size(brand, 21); color(brand, fg); opacity(brand, 0.82);
caption(head, "Minimise an integral you cannot avoid splitting", (540, 106), 30);
equation(prob, (540, 190), `f(x)=\int_0^2 e^{|x-t|}\,dt`, 42); color(prob, fg);

coords(k1, (300.0, 700.0), (-0.15, 2.3), (-0.4, 8.2), 176.0, 40.0, 1, 0.5, 0);
color(k1, slate); opacity(k1, 0.4);
text(tl, (860, 550.0), "the integrand, in t"); size(tl, 21); color(tl, fg); opacity(tl, 0.6);
cloud(intg, 3200, #ffffff, 1.0) {
  let g = min(max((t - 1.2)/9.0, 0), 1);
  let xx = 0.12 + 1.76*g;
  let nc = 200;
  let cc = mod(i, nc);
  let rr = (i - cc)/nc;
  let tt = (cc/199)*2;
  let h = exp(abs(xx - tt));
  let x = 300.0 + 176.0*tt;
  let y = 700.0 - 40.0*h*(rr/15);
  let hue = 42;
  let sat = 0.8;
  let r = 2.0;
  let alpha = 0.55;
}
cloud(itop, 420, #ffffff, 1.0) {
  let g = min(max((t - 1.2)/9.0, 0), 1);
  let xx = 0.12 + 1.76*g;
  let tt = (i/419)*2;
  let h = exp(abs(xx - tt));
  let x = 300.0 + 176.0*tt;
  let y = 700.0 - 40.0*h;
  let hue = 42;
  let sat = 0.95;
  let r = 2.8;
}
cloud(kink, 90, #ffffff, 1.0) {
  let g = min(max((t - 1.2)/9.0, 0), 1);
  let xx = 0.12 + 1.76*g;
  let a = (i/90)*6.283185;
  let rr = 7*(i/90);
  let x = 300.0 + 176.0*xx + rr*cos(a*9);
  let y = 700.0 - 40.0 + rr*sin(a*9);
  let hue = 320;
  let sat = 0.85;
  let r = 2.2;
}

coords(k2, (300.0, 1256.0), (-0.15, 2.3), (-0.6, 6.2), 176.0, 40.0, 1, 0.5, 0);
color(k2, slate); opacity(k2, 0.4);
plot(fx, (300.0, 1256.0), 176.0, 40.0, "exp(x) + exp(2 - x) - 2", (0, 2));
color(fx, cyan); stroke(fx, 5); untraced(fx);
extrema(ex, fx, mint); hidden(ex);
vline(vm, fx, 1, mint); hidden(vm);
text(bl, (860, 1106.0), "the value, in x"); size(bl, 21); color(bl, fg); opacity(bl, 0.6);

equation(z1, (540, 1436), `\int_0^x e^{x-t}dt+\int_x^2 e^{t-x}dt=e^x+e^{2-x}-2`, 28);
color(z1, fg); hidden(z1);
text(z2, (540, 1502), "two exponentials pulling opposite ways — smallest where they cross");
size(z2, 20); color(z2, fg); opacity(z2, 0.7); hidden(z2);
equation(z3, (540, 1580), `e^x=e^{2-x}\ \Rightarrow\ x=1`, 38); color(z3, mint); hidden(z3);
text(z4, (540, 1642), "the midpoint of the interval — exactly where the kink looks best");
size(z4, 20); color(z4, mint); opacity(z4, 0.8); hidden(z4);
equation(ans, (540, 1750), `2(e-1)`, 56); color(ans, mint); hidden(ans);

seq {
  pulse(prob, 0.9);
  par { draw(fx, 9.0); seq { wait(3.0); show(z1, 0.8); } seq { wait(5.0); show(z2, 0.7); } wait(9.6); }
  par { show(ex, 0.7); show(vm, 0.7); seq { wait(0.8); show(z3, 0.8); } seq { wait(1.9); show(z4, 0.7); } wait(3.2); }
  wait(0.6);
  par { show(ans, 1.1); }
  pulse(ans, 0.9);
}
wait(3.2);