定積分の微分
【問題】
$$
\int_\frac{\pi}{2}^x (x-t) f(t) dt =\sin x - a
$$
について $ f(x) $ を求めよ。
【解説】
$$ \int_\frac{\pi}{2}^x (x-t) f(t) dt =\sin x - a $$
$$ \int_\frac{\pi}{2}^x x f(t) dt - \int_\frac{\pi}{2}^x t f(t) dt =\sin x - a \text{(左辺の展開)} $$
$$ \left( x \cdot \int_\frac{\pi}{2}^x f(t) dt \right)^{\prime} - \left(\int_\frac{\pi}{2}^x t f(t) dt \right)^{\prime} =\left( \sin x - a\right)^{\prime} \text{(両辺を微分)} $$
$$ x^{\prime} \left( \int_\frac{\pi}{2}^x f(t) dt \right) + x \left( \int_\frac{\pi}{2}^x f(t) dt \right) ^{\prime} - x f(x) =\cos x \text{(積の微分に注意)} $$
$$ 1 \cdot \left( \int_\frac{\pi}{2}^x f(t) dt \right) + x f(x) - x f(x) =\cos x $$
$$ \int_\frac{\pi}{2}^x f(t) dt =\cos x $$
$$ f(x) =-\sin x \text{(両辺を微分)} $$
$$ \int_\frac{\pi}{2}^x x f(t) dt - \int_\frac{\pi}{2}^x t f(t) dt =\sin x - a \text{(左辺の展開)} $$
$$ \left( x \cdot \int_\frac{\pi}{2}^x f(t) dt \right)^{\prime} - \left(\int_\frac{\pi}{2}^x t f(t) dt \right)^{\prime} =\left( \sin x - a\right)^{\prime} \text{(両辺を微分)} $$
$$ x^{\prime} \left( \int_\frac{\pi}{2}^x f(t) dt \right) + x \left( \int_\frac{\pi}{2}^x f(t) dt \right) ^{\prime} - x f(x) =\cos x \text{(積の微分に注意)} $$
$$ 1 \cdot \left( \int_\frac{\pi}{2}^x f(t) dt \right) + x f(x) - x f(x) =\cos x $$
$$ \int_\frac{\pi}{2}^x f(t) dt =\cos x $$
$$ f(x) =-\sin x \text{(両辺を微分)} $$
質問と回答