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(A) Given $\int_{a}^{b} f(x) dx = \int_{a}^{b} g(x) dx$,we can write $\int_{a}^{b} (f(x) - g(x)) dx = 0$. Splitting the integral into three parts base

Vedclass · Accepted Answer

$\int_{\alpha}^{\beta} (g(x) - f(x)) dx = \int_{a}^{\alpha} (f(x) - g(x)) dx + \int_{\beta}^{b} (f(x) - g(x)) dx$

Vedclass · Answer

(A) Given $\int_{a}^{b} f(x) dx = \int_{a}^{b} g(x) dx$,we can write $\int_{a}^{b} (f(x) - g(x)) dx = 0$. Splitting the integral into three parts based on the intersection points $\alpha$ and $\beta$: $\int_{a}^{\alpha} (f(x) - g(x)) dx + \int_{\alpha}^{\beta} (f(x) - g(x)) dx + \int_{\beta}^{b} (f(x) - g(x)) dx = 0$. From the graph,in the interval $(a, \alpha)$,$f(x) > g(x)$,so the area $A_1 = \int_{a}^{\alpha} (f(x) - g(x)) dx$. In the interval $(\alpha, \beta)$,$g(x) > f(x)$,so the area $A_2 = \int_{\alpha}^{\beta} (g(x) - f(x)) dx = -\int_{\alpha}^{\beta} (f(x) - g(x)) dx$. In the interval $(\beta, b)$,$f(x) > g(x)$,so the area $A_3 = \int_{\beta}^{b} (f(x) - g(x)) dx$. Substituting these into the integral equation: $A_1 - A_2 + A_3 = 0$,which implies $A_2 = A_1 + A_3$. Thus,$\int_{\alpha}^{\beta} (g(x) - f(x)) dx = \int_{a}^{\alpha} (f(x) - g(x)) dx + \int_{\beta}^{b} (f(x) - g(x)) dx$.

Two differentiable functions $f(x)$ and $g(x)$ are such that $f''(x) > 0$ and $g''(x) < 0$ for all $x \in (a,b)$ and $\int_{a}^{b} f(x) dx = \int_{a}^{b} g(x) dx$. If $f(x) = g(x)$ for $x = \alpha, \beta \in (a,b)$ $(\alpha < \beta)$,then:

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