Let $$I = \int_a^b (x^4 - 2x^2)dx$$. If $$I$$ is minimum then the ordered pair $$(a, b)$$ is:

We have to compare the values of the definite integral

$$I=\int_a^b\,(x^4-2x^2)\,dx$$

for the four ordered pairs given in the options. The integrand is a polynomial, so first we find its antiderivative. Using the power‐rule formula

$$\int x^n\,dx=\frac{x^{\,n+1}}{n+1}+C,$$

we obtain

$$\int (x^4-2x^2)\,dx = \frac{x^5}{5}-\frac{2x^3}{3}+C.$$

Let us denote this antiderivative by

$$F(x)=\frac{x^5}{5}-\frac{2x^3}{3}.$$

For any limits $$a$$ and $$b$$, the definite integral is

$$I=F(b)-F(a).$$

Now we evaluate $$F(x)$$ at the two special numbers that appear in every option, namely $$x=0$$ and $$x=\pm\sqrt{2}\;.$$ First we compute

$$F(0)=\frac{0^5}{5}-\frac{2\cdot0^3}{3}=0.$$

Next we put $$x=\sqrt{2}:$$

$$F(\sqrt{2})=\frac{(\sqrt{2})^5}{5}-\frac{2(\sqrt{2})^{3}}{3}.$$

Because $$(\sqrt{2})^5=2^{5/2}=4\sqrt{2}$$ and $$(\sqrt{2})^3=2\sqrt{2},$$ we get

$$F(\sqrt{2})=\frac{4\sqrt{2}}{5}-\frac{2\cdot2\sqrt{2}}{3} =\frac{4\sqrt{2}}{5}-\frac{4\sqrt{2}}{3} =4\sqrt{2}\left(\frac15-\frac13\right) =4\sqrt{2}\left(\frac{3-5}{15}\right) =-\frac{8\sqrt{2}}{15}.$$

Putting $$x=-\sqrt{2}:$$

$$F(-\sqrt{2})=\frac{(-\sqrt{2})^5}{5}-\frac{2(-\sqrt{2})^{3}}{3} =\frac{-4\sqrt{2}}{5}-\frac{2(-2\sqrt{2})}{3} =\frac{-4\sqrt{2}}{5}+\frac{4\sqrt{2}}{3} =4\sqrt{2}\left(-\frac15+\frac13\right) =4\sqrt{2}\left(\frac{-3+5}{15}\right) =\frac{8\sqrt{2}}{15}.$$

With these three values of $$F(x)$$ we can now calculate the integral for each option.

Option A: $$(a,b)=\bigl(0,\sqrt{2}\bigr)$$

$$I_A=F(\sqrt{2})-F(0)=\left(-\frac{8\sqrt{2}}{15}\right)-0=-\frac{8\sqrt{2}}{15}.$$

Option B: $$(a,b)=\bigl(\sqrt{2},-\sqrt{2}\bigr)$$

$$I_B=F(-\sqrt{2})-F(\sqrt{2}) =\frac{8\sqrt{2}}{15}-\left(-\frac{8\sqrt{2}}{15}\right) =\frac{16\sqrt{2}}{15}.$$

Option C: $$(a,b)=\bigl(-\sqrt{2},0\bigr)$$

$$I_C=F(0)-F(-\sqrt{2})=0-\frac{8\sqrt{2}}{15}=-\frac{8\sqrt{2}}{15}.$$

Option D: $$(a,b)=\bigl(-\sqrt{2},\sqrt{2}\bigr)$$

$$I_D=F(\sqrt{2})-F(-\sqrt{2}) =\left(-\frac{8\sqrt{2}}{15}\right)-\frac{8\sqrt{2}}{15} =-\frac{16\sqrt{2}}{15}.$$

Comparing the four results, we have

$$I_B=\frac{16\sqrt{2}}{15}\;,\qquad I_A=I_C=-\frac{8\sqrt{2}}{15}\;,\qquad I_D=-\frac{16\sqrt{2}}{15}.$$

The smallest (most negative) value is clearly

$$I_D=-\frac{16\sqrt{2}}{15},$$

which corresponds to the ordered pair $$(-\sqrt{2},\sqrt{2}).$$

Hence, the correct answer is Option D.