Applications of the indefinite integral

Geometric: a curve from its slope

• If a curve y = f(x) passes through a known point and the slope of its tangent at any point is given, integrate the slope to get y = ∫ (slope) dx, then substitute the point to find C.
• Example: dy/dx = 2x − 3 and the curve passes through (2, 5). y = x² − 3x + C; 5 = 4 − 6 + C → C = 7, so y = x² − 3x + 7.
• The slope of the tangent is dy/dx; if instead the normal is given, use dy/dx = −1/(normal slope).

Physical: motion along a straight line

• Acceleration a = dv/dt, so v = ∫ a dt; velocity v = ds/dt, so s = ∫ v dt — each integration needs one initial condition (initial velocity, initial position).
• Example: a body starts from rest with a = 6t − 4. v = 3t² − 4t, s = t³ − 2t²; at t = 3, v = 15 m/s and s = 9 m.
• Work done by a variable force: W = ∫ F ds; and the change of momentum is ∫ F dt (see the vector-functions lesson).

Worked example (curve through two conditions)

1. A curve has second derivative d²y/dx² = 12x − 2 and passes through (0, 1) with tangent slope 4 there. dy/dx = 6x² − 2x + C₁ → C₁ = 4. y = 2x³ − x² + 4x + C₂ → C₂ = 1. y = 2x³ − x² + 4x + 1.