Define \[ F(x)=\int_{\sin x}^{\cos x} e^{t^2+xt}\,dt. \] Compute \(F'(0)\).
Solution
Set \[ a(x)\coloneqq\sin x, \qquad b(x)\coloneqq\cos x, \qquad f(x,t)\coloneqq e^{t^2+xt}. \]
1. For every \(x\),\[ F'(x) =-\sin x\,e^{\cos^2x+x\cos x} -\cos x\,e^{\sin^2x+x\sin x} +\int_{\sin x}^{\cos x} t e^{t^2+xt}\,dt. \]
The functions \(a\), \(b\), and \(f\) are smooth, so the Leibniz rule gives \[ F'(x) =f(x,b(x))b'(x)-f(x,a(x))a'(x) +\int_{a(x)}^{b(x)}\frac{\partial f}{\partial x}(x,t)\,dt. \] Here \(b'(x)=-\sin x\), \(a'(x)=\cos x\), and \[ \frac{\partial f}{\partial x}(x,t)=t e^{t^2+xt}, \] which yields the stated formula.
2. \(F'(0)=\boxed{\dfrac{e-3}{2}}\).
Substituting \(x=0\) into step 1 gives \[ F'(0)=-1+\int_0^1 t e^{t^2}\,dt. \] With \(u=t^2\), this integral is \[ \int_0^1 t e^{t^2}\,dt =\frac12\int_0^1 e^u\,du =\frac{e-1}{2}. \] Therefore \[ F'(0)=-1+\frac{e-1}{2}=\frac{e-3}{2}. \]
3. Q.E.D.
Step 2 gives the requested value.