Voltage through space and time

We have $B = \sin(x^2 + t), \\ dA = y dx = x dx.$

$\lambda$ is a flux linkage, and $d\lambda = B da = x \sin(x^2 + t)\ dx \\ \rightarrow \lambda = \int_0^1 x \sin(x^2 + t)\ dx.$

Let $u = x^2 + t,$ then $\frac{du}{dx} = 2x \implies dx = \frac{du}{2x}.$ So, $\begin{aligned} \lambda &= \frac12 \int_t^{1+t} x \sin u \ \frac{du}{x} \\ &= \frac12 \int_t^{1+t} \sin u \ du \\ &= -\frac12 \cos u \Big|_t^{1+t} \\ &= \frac12[\cos t - \cos(1+t)]. \end{aligned}$

Resultant will be a sinusoid, so max and min have same magnitude.

Let $V = \frac{d\lambda}{dt}.$ To find max $V,$ take $\frac{dV}{dt} = \frac{d^2\lambda}{dt^2} = 0$ which leaves us with the same expression for $\lambda.$

Thus, $\cos t = \cos(1+t).$

Using $\cos \alpha = \cos (-\alpha),$ $t = -1 -t \implies t_{max} = -\frac12.$

Since signal is periodic, $\begin{aligned} t_{max} &= -\frac12 \pm 2\pi n, \qquad n=0,1,2, \ldots. \\ V &= \frac12 \big[-\sin t + \sin(1+t) \big] \\ V_{max} &= \frac12 \left[-\sin \left(-\frac12 \right) + \sin \left(\frac12 \right) \right] \approx \boxed{0.479}. \end{aligned}$

After obtaining flux the same way that @Steven Chase obtained

I Used the trigonometric identity cosC-cosD = 2sin(C+D/2)sin(D-C/2)

Using this and putting we get Flux as sin(0.5)sin(t+0.5)

Differentiate this to get induced emf sin(0.5)cos(t+0.5)

EMF will be max when cos(t+0.5) = 1

So Emf max = sin(0.5) = 0.479