This is simple(but tricky!)

Taking the inverse of a function can be viewed as rotating the graph about the line $y=x$ , i.e., interchanging the axes.

The graph of $f(x)$ looks somewhat like this (on finding $f'(x)$ , we see that it is positive for $x \geq 0$ ). --

graph

The area we need to find is that shaded, the area that $f(x)$ makes with the $y$ axis from $y=1$ to $y=2015$ .

On some observation of $f(x)$ , we see that it attains the value $2015$ at $x=1$ .

So, the required area can be written as the area enclosed by $f(x)$ with the $x$ axis from $0$ to $1$ subtracted from a rectangle with length $2015$ units and breadth $1$ unit.

Hence, $\begin{aligned} \displaystyle\int_{1}^{2015} f^{-1}(x) \ dx &=2015 \times 1-\displaystyle\int_{0}^{1} f(x) \ dx \\ & = 2015-\left(\dfrac{1}{2015}+\dfrac{2013}{\log {(2014)}}\right) \\ & \approx \boxed{1750.41} \end{aligned}$

We use substitution $\displaystyle x=f(u) , dx=f'(u) du$

The limits of integration will be then:

For $\displaystyle x=1 , u^{2014} +2014^u = 1 => u=0$ .

For $\displaystyle x=2015 , u^{2014} +2014^u = 2015 => u=1$ .

By the definition of the inverse function, $\displaystyle f^{-1}(f(u))=u$ .

Therefore,

$\displaystyle\int_1^{2015} \mathrm f^{-1}(x)\,\mathrm{d}x = \int_0^1 \mathrm uf'(u)\,\mathrm{d}u$

Integrating by parts:

$\displaystyle\int_0^1 \mathrm uf'(u)\,\mathrm{d}u =\left.uf(u)\right|_0^1 -\int_0^1 \mathrm f(u)\,\mathrm{d}u$

$\displaystyle = \left.u( u^{2014} +2014^u)\right|_0^1- \int_0^1 \mathrm u^{2014} +2014^u\,\mathrm{d}u$

$\displaystyle = \left.u(u^{2014} +2014^u)-\frac{u^{2015}}{2015} -\frac{2014^u}{\ln 2014}\right|_0^1$

$\displaystyle = 2015-\frac{1}{2015} -\frac{2014}{\ln 2014}-\frac{1}{\ln 2014} \approx \boxed{1750.4}$

Consider a visual proof of integration by parts:

Note it's obvious that $f(0)=1,f(1)=2015$ and thus $f^{-1}(1)=0,f^{-1}(2015)=1$ . It follows by the above proof that $f(1)\cdot f^{-1}(2015) - f(0)\cdot f^{-1}(1)=\int_0^1 f(x)~dx+\int_{f(0)}^{f(1)} f^{-1}(x)~dx$

Simplifying, we find $\begin{aligned}\int_1^{2015} f^{-1}(x)~dx&=1\cdot2015-\int_0^1(x^{2014}+2014^x)\ dx\\&=2014-\left[\frac1{2015}x^{2015}+\frac1{\log(2014)}2014^x\right]_0^1\\&=2014-\frac1{2015}-\frac{2013}{\log(2014)}\\&\approx1750.4\end{aligned}$

substitute x=f(y) , , by inspection we get limits as 0 to 1 (corresponding to 1 and 2015)

Then you get integral of yf'(y) dy from 0 to 1, take y as first function and f'(y) as second and use IBP,, Fortunately the original function is very easy to integrate,, and ultimately u get 1750.40