Integrate, add a constant, then substitute the given point to determine it. Example Find the curve with gradient function dydx=3x2+2 that passes through (1,6). y = (3x2+2),dx =x3+2x+C Substitute (1,6): 6 =13+2(1)+C C =3 y=x3+2x+3 Using a known derivative g (x)=h(x) kh(x), dx=kg(x)+C For a definite integral, _abh(x), dx=g(b)-g(a), where the integral exists. Second derivatives **Given** | **Integrate** | **Fixing the constants** dydx and a point on the curve | Once | Substitute the point to find the constant. d2ydx2, the gradient at a point and a point on the curve | Twice | The gradient fixes the first constant; the point fixes the second. Each integration introduces a new constant. Gradient information fixes the first; a point fixes the second. Disconnected domain intervals may have independent constants.

Integrate, add a constant, then substitute the given point to determine it.

Find the curve with gradient function \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=3x^2+2\) that passes through \((1,6)\).

\(y\)\({}=\int(3x^2+2)\,\mathrm{d}x\)\({}=x^3+2x+C\)

\(6\)\({}=1^3+2(1)+C\)\(C\)\({}=3\)

\(\mathrm{g}^{\prime}(x)=h(x)\quad\Longrightarrow\quad\int kh(x)\,\mathrm dx=kg(x)+C\)

For a definite integral, \(\int_a^bh(x)\,\mathrm dx=\mathrm{g}(b)-\mathrm{g}(a)\), where the integral exists.

Recovering a Curve from Its Derivatives

Recovering a Curve from Its Derivatives