Question

cho hình phẳng giới hạn bởi đồ thị hàm số \(y=4xe^x\), trục hoành và các đương thẳng x=1,x=6. Tính thể tích khối tròn xoay tạo thành khi cho hình phẳng đó quay quanh trục Ox

Answer 1

Thể tích khối tròn xoay tạo thành khi cho hình phẳng đó quay quanh trục Ox là

\(V=\pi\int_1^6\left(4xe^x\right)^2dx=16\pi\int_1^6x^2e^{2x}dx\)

Gọi \(I=\int x^2e^{2x}dx\)

Đặt \(\left\{{}\begin{matrix}u=x^2\\dv=e^{2x}dx\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}du=2xdx\\v=\dfrac{1}{2}e^{2x}\end{matrix}\right.\)

\(\Rightarrow I=\dfrac{1}{2}x^2e^{2x}-\int\dfrac{1}{2}e^{2x}2xdx=\dfrac{1}{2}x^2e^{2x}-\int e^{2x}xdx\)

Gọi \(K=\int e^{2x}xdx\)

Đặt \(\left\{{}\begin{matrix}u=x\\dv=e^{2x}dx\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}du=dx\\v=\dfrac{1}{2}e^{2x}\end{matrix}\right.\)

\(\Rightarrow K=\dfrac{1}{2}xe^{2x}-\int\dfrac{1}{2}e^{2x}dx=\dfrac{1}{2}xe^{2x}-\dfrac{1}{4}e^{2x}+C\)

\(\Rightarrow I=\dfrac{1}{2}x^2e^{2x}-\dfrac{1}{2}xe^{2x}+\dfrac{1}{4}e^{2x}+C\)

\(\Rightarrow\int_1^6x^2e^{2x}dx=\left(\dfrac{1}{2}x^2e^{2x}-\dfrac{1}{2}xe^{2x}+\dfrac{1}{4}e^{2x}\right)|^6_1\)

\(=\dfrac{36e^{12}}{2}-\dfrac{6e^{12}}{2}+\dfrac{e^{12}}{4}-\dfrac{e^2}{2}+\dfrac{e^2}{2}-\dfrac{e^2}{4}\)

\(=\dfrac{61e^{12}-e^2}{4}\)

\(\Rightarrow V=16\pi.\dfrac{61e^{12}-e^2}{4}=4\pi\left(61e^{12}-e^2\right)\)

Answer 2

Thể tích khối tròn xoay tạo thành khi cho hình phẳng đó quay quanh trục Ox là

\(V=\pi\int_1^6\left(4xe^x\right)^2dx=16\pi\int_1^6x^2e^{2x}dx\)

Gọi \(I=\int x^2e^{2x}dx\)

Đặt \(\left\{{}\begin{matrix}u=x^2\\dv=e^{2x}dx\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}du=2xdx\\v=\dfrac{1}{2}e^{2x}\end{matrix}\right.\)

\(\Rightarrow I=\dfrac{1}{2}x^2e^{2x}-\int\dfrac{1}{2}e^{2x}2xdx=\dfrac{1}{2}x^2e^{2x}-\int e^{2x}xdx\)

Gọi \(K=\int e^{2x}xdx\)

Đặt \(\left\{{}\begin{matrix}u=x\\dv=e^{2x}dx\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}du=dx\\v=\dfrac{1}{2}e^{2x}\end{matrix}\right.\)

\(\Rightarrow K=\dfrac{1}{2}xe^{2x}-\int\dfrac{1}{2}e^{2x}dx=\dfrac{1}{2}xe^{2x}-\dfrac{1}{4}e^{2x}+C\)

\(\Rightarrow I=\dfrac{1}{2}x^2e^{2x}-\dfrac{1}{2}xe^{2x}+\dfrac{1}{4}e^{2x}+C\)

\(\Rightarrow\int_1^6x^2e^{2x}dx=\left(\dfrac{1}{2}x^2e^{2x}-\dfrac{1}{2}xe^{2x}+\dfrac{1}{4}e^{2x}\right)|^6_1\)

\(=\dfrac{36e^{12}}{2}-\dfrac{6e^{12}}{2}+\dfrac{e^{12}}{4}-\dfrac{e^2}{2}+\dfrac{e^2}{2}-\dfrac{e^2}{4}\)

\(=\dfrac{61e^{12}-e^2}{4}\)

\(\Rightarrow V=16\pi.\dfrac{61e^{12}-e^2}{4}=4\pi\left(61e^{12}-e^2\right)\)

Answer 3

Thể tích khối tròn xoay tạo thành khi cho hình phẳng đó quay quanh trục Ox là

\(V=\pi\int_1^6\left(4xe^x\right)^2dx=16\pi\int_1^6x^2e^{2x}dx\)

Gọi \(I=\int x^2e^{2x}dx\)

Đặt \(\left\{{}\begin{matrix}u=x^2\\dv=e^{2x}dx\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}du=2xdx\\v=\dfrac{1}{2}e^{2x}\end{matrix}\right.\)

\(\Rightarrow I=\dfrac{1}{2}x^2e^{2x}-\int\dfrac{1}{2}e^{2x}2xdx=\dfrac{1}{2}x^2e^{2x}-\int e^{2x}xdx\)

Gọi \(K=\int e^{2x}xdx\)

Đặt \(\left\{{}\begin{matrix}u=x\\dv=e^{2x}dx\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}du=dx\\v=\dfrac{1}{2}e^{2x}\end{matrix}\right.\)

\(\Rightarrow K=\dfrac{1}{2}xe^{2x}-\int\dfrac{1}{2}e^{2x}dx=\dfrac{1}{2}xe^{2x}-\dfrac{1}{4}e^{2x}+C\)

\(\Rightarrow I=\dfrac{1}{2}x^2e^{2x}-\dfrac{1}{2}xe^{2x}+\dfrac{1}{4}e^{2x}+C\)

\(\Rightarrow\int_1^6x^2e^{2x}dx=\left(\dfrac{1}{2}x^2e^{2x}-\dfrac{1}{2}xe^{2x}+\dfrac{1}{4}e^{2x}\right)|^6_1\)

\(=\dfrac{36e^{12}}{2}-\dfrac{6e^{12}}{2}+\dfrac{e^{12}}{4}-\dfrac{e^2}{2}+\dfrac{e^2}{2}-\dfrac{e^2}{4}\)

\(=\dfrac{61e^{12}-e^2}{4}\)

\(\Rightarrow V=16\pi.\dfrac{61e^{12}-e^2}{4}=4\pi\left(61e^{12}-e^2\right)\)