Answer
see the details below.
Work Step by Step
We have
$$\langle x^2, 1\rangle=\int_{-1}^1 x^2 d x=\left[ \frac{ 1}{3 } x^3\right]_{-1}^1=\frac{ 1}{3 }+\frac{ 1}{3 } =\frac{ 2}{3 }.$$
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