For f(x) = x^4 + |x|,let I_1 = int_{0}^{pi} f | vedclass

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(C) Given $f(x) = x^4 + |x|$. Since $f(-x) = (-x)^4 + |-x| = x^4 + |x| = f(x)$,$f(x)$ is an even function.$I_1 = \int_{0}^{\pi} f(\cos x) dx$.Using th

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(C) Given $f(x) = x^4 + |x|$. Since $f(-x) = (-x)^4 + |-x| = x^4 + |x| = f(x)$,$f(x)$ is an even function.$I_1 = \int_{0}^{\pi} f(\cos x) dx$.Using the property $\int_{0}^{a} g(x) dx = \int_{0}^{a} g(a-x) dx$,we get:$I_1 = \int_{0}^{\pi} f(\cos(\pi - x)) dx = \int_{0}^{\pi} f(-\cos x) dx$.Since $f(x)$ is an even function,$f(-\cos x) = f(\cos x)$.Thus,$I_1 = \int_{0}^{\pi} f(\cos x) dx = 2 \int_{0}^{\frac{\pi}{2}} f(\cos x) dx$.Now,consider $I_2 = \int_{0}^{\frac{\pi}{2}} f(\sin x) dx$. Using the property $\int_{0}^{a} g(x) dx = \int_{0}^{a} g(a-x) dx$:$I_2 = \int_{0}^{\frac{\pi}{2}} f(\sin(\frac{\pi}{2} - x)) dx = \int_{0}^{\frac{\pi}{2}} f(\cos x) dx$.Therefore,$I_1 = 2 I_2$,which implies $\frac{I_1}{I_2} = 2$.

For $f(x) = x^4 + |x|$,let $I_1 = \int_{0}^{\pi} f(\cos x) dx$ and $I_2 = \int_{0}^{\frac{\pi}{2}} f(\sin x) dx$. Then $\frac{I_1}{I_2}$ is equal to

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