To find the numerical value of int_{-2}^{2} ( | vedclass

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(D) Let the given integral be $I = \int_{-2}^{2} (px^2 + qx + s) \, dx$.We can split this into three integrals:$I = p \int_{-2}^{2} x^2 \, dx + q \int

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$p$ and $s$

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(D) Let the given integral be $I = \int_{-2}^{2} (px^2 + qx + s) \, dx$.We can split this into three integrals:$I = p \int_{-2}^{2} x^2 \, dx + q \int_{-2}^{2} x \, dx + s \int_{-2}^{2} 1 \, dx$.Since $f(x) = x$ is an odd function,$\int_{-2}^{2} x \, dx = 0$.Since $f(x) = x^2$ and $f(x) = 1$ are even functions,we can use the property $\int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx$.Thus,$I = 2p \int_{0}^{2} x^2 \, dx + 0 + 2s \int_{0}^{2} 1 \, dx$.$I = 2p \left[ \frac{x^3}{3} \right]_{0}^{2} + 2s [x]_{0}^{2}$.$I = 2p \left( \frac{8}{3} \right) + 2s(2) = \frac{16p}{3} + 4s$.Therefore,to find the numerical value of $I$,it is necessary to know the values of $p$ and $s$.

To find the numerical value of $\int_{-2}^{2} (px^2 + qx + s) \, dx$,it is necessary to know the values of which constants?

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