Let $$f, g: [0, \infty) \rightarrow R$$ be two functions defined by $$f(x) = \int_{-x}^{x}(|t| - t^2)e^{-t^2}dt$$ and $$g(x) = \int_{0}^{x^2}t^{1/2}e^{-t}dt$$. Then the value of $$9(f(\sqrt{\log_e 9}) + g(\sqrt{\log_e 9}))$$ is equal to

Since $$(|t| - t^2)\,e^{-t^2}$$ is an even function of $$t$$ (replacing $$t$$ by $$-t$$ leaves it unchanged), the integral of $$f$$ over the symmetric interval $$[-x,x]$$ becomes:

$$f(x) = 2\int_0^x (t - t^2)\,e^{-t^2}\,dt = 2\int_0^x t\,e^{-t^2}\,dt - 2\int_0^x t^2\,e^{-t^2}\,dt.$$

The first integral evaluates directly via $$u = t^2$$:

$$2\int_0^x t\,e^{-t^2}\,dt = \bigl[-e^{-t^2}\bigr]_0^x = 1 - e^{-x^2}.$$

$$f(x) = 1 - e^{-x^2} - 2\int_0^x t^2\,e^{-t^2}\,dt. \quad\cdots(1)$$

For $$g(x) = \int_0^{x^2}\!\sqrt{t}\;e^{-t}\,dt$$, substitute $$t = u^2$$, so $$dt = 2u\,du$$ and $$\sqrt{t} = u$$. When $$t = 0$$, $$u = 0$$; when $$t = x^2$$, $$u = x$$. Then:

$$g(x) = \int_0^x u \cdot e^{-u^2}\cdot 2u\,du = 2\int_0^x u^2\,e^{-u^2}\,du. \quad\cdots(2)$$

Adding (1) and (2), the integral terms cancel exactly:

$$f(x) + g(x) = 1 - e^{-x^2} - 2\int_0^x t^2 e^{-t^2}\,dt + 2\int_0^x u^2 e^{-u^2}\,du = 1 - e^{-x^2}.$$

Substituting $$x = \sqrt{\log_e 9}$$:

$$e^{-x^2} = e^{-\log_e 9} = \frac{1}{9}.$$

$$f\!\left(\sqrt{\log_e 9}\right) + g\!\left(\sqrt{\log_e 9}\right) = 1 - \frac{1}{9} = \frac{8}{9}.$$

Finally, $$9 \times \frac{8}{9} = 8$$.