次の定積分を計算します。 $\int_0^{\sqrt{e}} [\frac{1}{4}x^4 - (e \log x)^2] dx$解析学定積分部分積分対数関数2025/5/41. 問題の内容次の定積分を計算します。∫0e[14x4−(elogx)2]dx\int_0^{\sqrt{e}} [\frac{1}{4}x^4 - (e \log x)^2] dx∫0e[41x4−(elogx)2]dx2. 解き方の手順まず、積分を分解します。∫0e[14x4−(elogx)2]dx=∫0e14x4dx−∫0e(elogx)2dx\int_0^{\sqrt{e}} [\frac{1}{4}x^4 - (e \log x)^2] dx = \int_0^{\sqrt{e}} \frac{1}{4}x^4 dx - \int_0^{\sqrt{e}} (e \log x)^2 dx∫0e[41x4−(elogx)2]dx=∫0e41x4dx−∫0e(elogx)2dx次に、それぞれの積分を計算します。∫0e14x4dx=14∫0ex4dx=14[15x5]0e=120(e)5=120(e1/2)5=120e5/2=e5/220\int_0^{\sqrt{e}} \frac{1}{4}x^4 dx = \frac{1}{4} \int_0^{\sqrt{e}} x^4 dx = \frac{1}{4} [\frac{1}{5}x^5]_0^{\sqrt{e}} = \frac{1}{20} (\sqrt{e})^5 = \frac{1}{20} (e^{1/2})^5 = \frac{1}{20} e^{5/2} = \frac{e^{5/2}}{20}∫0e41x4dx=41∫0ex4dx=41[51x5]0e=201(e)5=201(e1/2)5=201e5/2=20e5/2次に、∫0e(elogx)2dx\int_0^{\sqrt{e}} (e \log x)^2 dx∫0e(elogx)2dx を計算します。∫0e(elogx)2dx=e2∫0e(logx)2dx\int_0^{\sqrt{e}} (e \log x)^2 dx = e^2 \int_0^{\sqrt{e}} (\log x)^2 dx∫0e(elogx)2dx=e2∫0e(logx)2dx∫(logx)2dx\int (\log x)^2 dx∫(logx)2dx を部分積分で計算します。u=(logx)2,dv=dxu = (\log x)^2, dv = dxu=(logx)2,dv=dx とすると、du=2(logx)1xdx,v=xdu = 2 (\log x) \frac{1}{x} dx, v = xdu=2(logx)x1dx,v=x∫(logx)2dx=x(logx)2−∫x⋅2(logx)1xdx=x(logx)2−2∫logxdx\int (\log x)^2 dx = x (\log x)^2 - \int x \cdot 2 (\log x) \frac{1}{x} dx = x (\log x)^2 - 2 \int \log x dx∫(logx)2dx=x(logx)2−∫x⋅2(logx)x1dx=x(logx)2−2∫logxdx∫logxdx\int \log x dx∫logxdx を部分積分で計算します。u=logx,dv=dxu = \log x, dv = dxu=logx,dv=dx とすると、du=1xdx,v=xdu = \frac{1}{x} dx, v = xdu=x1dx,v=x∫logxdx=xlogx−∫x⋅1xdx=xlogx−∫dx=xlogx−x\int \log x dx = x \log x - \int x \cdot \frac{1}{x} dx = x \log x - \int dx = x \log x - x∫logxdx=xlogx−∫x⋅x1dx=xlogx−∫dx=xlogx−xしたがって、∫(logx)2dx=x(logx)2−2(xlogx−x)=x(logx)2−2xlogx+2x\int (\log x)^2 dx = x (\log x)^2 - 2(x \log x - x) = x (\log x)^2 - 2x \log x + 2x∫(logx)2dx=x(logx)2−2(xlogx−x)=x(logx)2−2xlogx+2x∫0e(logx)2dx=[x(logx)2−2xlogx+2x]0e\int_0^{\sqrt{e}} (\log x)^2 dx = [x (\log x)^2 - 2x \log x + 2x]_0^{\sqrt{e}}∫0e(logx)2dx=[x(logx)2−2xlogx+2x]0e=[e(loge)2−2eloge+2e]−limx→0[x(logx)2−2xlogx+2x]= [\sqrt{e} (\log \sqrt{e})^2 - 2 \sqrt{e} \log \sqrt{e} + 2\sqrt{e}] - \lim_{x \to 0} [x (\log x)^2 - 2x \log x + 2x]=[e(loge)2−2eloge+2e]−limx→0[x(logx)2−2xlogx+2x]loge=loge1/2=12loge=12\log \sqrt{e} = \log e^{1/2} = \frac{1}{2} \log e = \frac{1}{2}loge=loge1/2=21loge=21e(12)2−2e(12)+2e=e4−e+2e=e4+e=5e4\sqrt{e} (\frac{1}{2})^2 - 2 \sqrt{e} (\frac{1}{2}) + 2 \sqrt{e} = \frac{\sqrt{e}}{4} - \sqrt{e} + 2\sqrt{e} = \frac{\sqrt{e}}{4} + \sqrt{e} = \frac{5\sqrt{e}}{4}e(21)2−2e(21)+2e=4e−e+2e=4e+e=45elimx→0x(logx)2=0,limx→0xlogx=0,limx→0x=0\lim_{x \to 0} x (\log x)^2 = 0, \lim_{x \to 0} x \log x = 0, \lim_{x \to 0} x = 0limx→0x(logx)2=0,limx→0xlogx=0,limx→0x=0∫0e(logx)2dx=5e4\int_0^{\sqrt{e}} (\log x)^2 dx = \frac{5\sqrt{e}}{4}∫0e(logx)2dx=45e∫0e(elogx)2dx=e2∫0e(logx)2dx=e2⋅5e4=5e5/24\int_0^{\sqrt{e}} (e \log x)^2 dx = e^2 \int_0^{\sqrt{e}} (\log x)^2 dx = e^2 \cdot \frac{5\sqrt{e}}{4} = \frac{5e^{5/2}}{4}∫0e(elogx)2dx=e2∫0e(logx)2dx=e2⋅45e=45e5/2∫0e[14x4−(elogx)2]dx=e5/220−5e5/24=e5/220−25e5/220=−24e5/220=−6e5/25\int_0^{\sqrt{e}} [\frac{1}{4}x^4 - (e \log x)^2] dx = \frac{e^{5/2}}{20} - \frac{5e^{5/2}}{4} = \frac{e^{5/2}}{20} - \frac{25e^{5/2}}{20} = \frac{-24e^{5/2}}{20} = \frac{-6e^{5/2}}{5}∫0e[41x4−(elogx)2]dx=20e5/2−45e5/2=20e5/2−2025e5/2=20−24e5/2=5−6e5/23. 最終的な答え−6e5/25-\frac{6e^{5/2}}{5}−56e5/2