Bài 3b: Luyện kĩ năng: Nguyên hàm, tích phân và ứng dụng

1.

\(V=\pi \int ^4_1[x^{\frac{1}{2}}e^{\frac{x}{2}}]^2dx=\pi \int ^4_1(xe^x)dx\)

\(=\pi \int ^4_1xd(e^x)=\pi (|^4_1xe^x-\int ^4_1e^xdx)\)

\(=\pi |^4_1(xe^x-e^x)=\pi (3e^4)=3\pi e^4\) 

2.

\(V=\pi \int ^1_0(x\sqrt{\ln (x^3+1)})^2dx=\pi \int ^1_0x^2\ln (x^3+1)dx\)

\(=\frac{1}{3}\pi \int ^1_0\ln (x^3+1)d(x^3+1)\)

\(=\frac{1}{3}\pi \int ^2_1ln tdt=\frac{1}{3}\pi (|^2_1t\ln t-\int ^2_1td(\ln t))\)

\(=\frac{1}{3}\pi (|^2_1t\ln t-\int ^2_1dt)=\frac{1}{3}\pi |^2_1(t\ln t-t)=\frac{1}{3}\pi (2\ln 2-1)\)

Lời giải:

Ta có : \(P=\int x^7\sqrt{x^4+4}dx=\int x^4\sqrt{x^4+4}(x^3dx)\)

\(=\int x^4\sqrt{x^4+4}\frac{d(x^4+4)}{4}=\frac{1}{4}\int x^4\sqrt{x^4+4}d(x^4+4)\)

Đặt \(\sqrt{x^4+4}=t\Rightarrow x^4=t^2-4\)

Khi đó:

\(P=\frac{1}{4}\int (t^2-4)td(t^2)=\frac{1}{2}\int (t^2-4)t^2dt\)

\(=\frac{1}{2}\int t^4dt-2\int t^2dt=\frac{t^5}{10}-\frac{2t^3}{3}+c\)

\(=\frac{1}{10}\sqrt{(x^4+4)^5}-\frac{2}{3}\sqrt{(x^4+4)^3}+c\)

Vậy \(m=\frac{1}{10}; n=\frac{2}{3}\Rightarrow mn=\frac{1}{15}\)

Lời giải:

Theo nhị thức New-ton:

\((x+1)^{2n}=C^{0}_{2n}+C^{1}_{2n}x+C^2_{2n}x^2+...+C^{2n}_{2n}x^{2n}\)

\((x-1)^n=C^0_{2n}-C^1_{2n}x+C^2_{2n}x^2-.....-C^{2n-1}_{2n}x^{2n-1}+C^{2n}_{2n}x^{2n}\)

Trừ theo vế ta có:

\(\frac{(x+1)^{2n}-(x-1)^{2n}}{2}=C^1_{2n}x+C^3_{2n}x^3+...+C^{2n-1}_{2n}x^{2n-1}\)

\(\Rightarrow \int ^{1}_{0}\frac{(x+1)^{2n}-(x-1)^{2n}}{2}dx=\int ^{1}_{0}(C^1_{2n}x+C^3_{2n}x^3+...+C^{2n-1}_{2n}x^{2n-1})dx\)

Xét vế trái:

\(\text{VT}=\frac{1}{2}\int ^{1}_{0}(x+1)^{2n}d(x+1)-\frac{1}{2}\int ^{1}_{0}(x-1)^{2n}d(x-1)\)

\(=\left.\begin{matrix} 1\\ 0\end{matrix}\right|\frac{1}{2}\left ( \frac{(x+1)^{2n+1}-(x-1)^{2n+1}}{2n+1} \right )=\frac{2^{2n}-1}{2n+1}\)

Xét vế phải:

\(\text{VP}=\left.\begin{matrix} 1\\ 0\end{matrix}\right|\left ( \frac{C^{1}_{2n}x^2}{2}+\frac{C^{3}_{2n}x^4}{4}+....+\frac{C^{2n-1}_{2n}x^{2n}}{2n} \right )=\frac{1}{2}C^{1}_{2n}+\frac{1}{4}C^3_{2n}+...+\frac{1}{2n}C^{2n-1}_{2n}\)

Vậy \(A=\frac{2^{2n}-1}{2n+1}\)