If a curve $$y = f(x)$$ passes through the point (1, 2) and satisfies $$x\frac{dy}{dx} + y = bx^4$$, then for what value of $$b$$, $$\int_1^2 f(x)dx = \frac{62}{5}$$?

The curve $$y = f(x)$$ passes through $$(1, 2)$$ and satisfies the differential equation $$x\frac{dy}{dx} + y = bx^4$$.

The left side can be written as $$\frac{d}{dx}(xy) = bx^4$$. Integrating both sides with respect to $$x$$, we get $$xy = \frac{bx^5}{5} + C$$.

Using the initial condition $$f(1) = 2$$: $$1 \cdot 2 = \frac{b}{5} + C$$, so $$C = 2 - \frac{b}{5}$$.

Therefore, $$xy = \frac{bx^5}{5} + 2 - \frac{b}{5}$$, which gives $$f(x) = \frac{bx^4}{5} + \frac{2 - \frac{b}{5}}{x}$$.

Now we compute $$\int_1^2 f(x)\,dx = \int_1^2 \left[\frac{bx^4}{5} + \frac{(2 - \frac{b}{5})}{x}\right]dx = \left[\frac{bx^5}{25} + \left(2 - \frac{b}{5}\right)\ln x\right]_1^2$$.

Evaluating: $$= \frac{32b}{25} - \frac{b}{25} + \left(2 - \frac{b}{5}\right)\ln 2 = \frac{31b}{25} + \left(2 - \frac{b}{5}\right)\ln 2$$.

Setting this equal to $$\frac{62}{5}$$: $$\frac{31b}{25} + \left(2 - \frac{b}{5}\right)\ln 2 = \frac{62}{5}$$. For this to hold without a logarithmic term, we need $$2 - \frac{b}{5} = 0$$, giving $$b = 10$$. Then $$\frac{31 \cdot 10}{25} = \frac{310}{25} = \frac{62}{5}$$, which confirms the equation.

Therefore, $$b = 10$$.