【高校数学Ⅲ】5-2-3 定積分と導関数｜問題集

\(f'(x)=\sin x\)

\(f'(x)=x\log x\)

\(f'(x)=\sin x\log(x^2+1)\)

\(\displaystyle =\int_{\frac{\pi}{2}}^x (x\cos t-t\cos t)dt\)
\(\displaystyle f'(x)=(\int_{\frac{\pi}{2}}^x x\cos tdt)'-x\cos x\)
\(\displaystyle =\int_{\frac{\pi}{2}}^x \cos tdt+x(\int_{\frac{\pi}{2}}^x \cos tdt)'-x\cos x\)
\(\displaystyle =\sin x-1+x\cos x-x\cos x\)
\(\displaystyle =\sin x-1\)

\(\displaystyle f(x)=\int_0^x e^tdt-\int_0^x te^tdt\)
\(\displaystyle f'(x)=\int_0^x e^tdt+xe^x-xe^x=[e^t]_0^x=e^x-1\)
\(f''(x)=e^x\)

\(\displaystyle a=\int_0^1 tf(t)dt\)とおくと、
\(f(x)=e^x+a\)
\(f(t)=e^t+a\)
\(\displaystyle a=\int_0^1 t(e^t+a)dt\)
\(\displaystyle a=\int_0^1 te^tdt+a\int_0^1 tdt\)
\(\displaystyle a=\int_0^1 te^tdt+a\left[\frac{1}{2}t^2\right]_0^1\)
\(\displaystyle a=[te^t]_0^1-\int_0^1 e^tdt+\frac{1}{2}a\)
\(\displaystyle \frac{1}{2}a=e-(e-1)\)
\(\displaystyle a=2\)
よって、
\(\displaystyle f(x)=e^x+2\)

\(\displaystyle a=\int_0^1 f(t)e^tdt\)とおくと、
\(f(x)=x+a\)
\(f(t)=t+a\)
\(\displaystyle a=\int_0^1 (t+a)e^tdt\)
\(\displaystyle a=[(t+a)e^t]_0^1-\int_0^1 e^tdt\)
\(\displaystyle a=(1+a)e-a-(e-1)\)
\(\displaystyle a=\frac{1}{2-e}\)
よって、
\(\displaystyle f(x)=x+\frac{1}{2-e}\)