Let f(x) and g(x) be positive differentiable functions on R such that g(x) = integral from -infinity to x of f(t) dt, with f(0) = 1 and g(0) = 4. The tangents to y = f(x) and y = g(x) at points with the same x-coordinate always meet on the x-axis. Find the value of the integral from -1 to

Question

Let f(x) and g(x) be positive differentiable functions on R such that g(x) = integral from -infinity to x of f(t) dt, with f(0) = 1 and g(0) = 4. The tangents to y = f(x) and y = g(x) at points with the same x-coordinate always meet on the x-axis. Find the value of the integral from -1 to 1 of (x² + 4) / (4 + g(x)) dx.

Accepted Answer

integral from -1 to 1 of (x²+4)/(4+g(x)) dx = 13/12. From the tangent intersection condition: f² = g*f'. Since g'=f, we have f*df/dg = f²/g => df/f = dg/g => f = K*g. Using g'=f=Kg: g(x) = g(0)*e^(Kx) = 4*e^(Kx). Then f = 4K*e^(Kx). f(0)=4K=1 => K=1/4. So f(x)=e^(x/4), g(x)=4*e^(x/4). Integral I = integral from -1 to 1 of (x²+4)/(4+4*e^(x/4)) dx = (1/4)*integral of (x²+4)/(1+e^(x/4)) dx. Use symmetry: let J = integral₋₁¹ (x²+4)/(1+e^(x/4))dx. J + J(x->-x) = integral₋₁¹(x²+4)dx = [x³/3+4x]₋₁¹ = (1/3+4)-(-1/3-4)=2/3+8=26/3. So 2J=26/3 => J=13/3. I = J/4 = 13/12.

Solution

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