Trang chủ » Hỏi đáp » Môn Toán cho f(x) liên tục trên R thỏa f(2x) – xf($x^{2}$) = $x^{3}$. khi đó $\int\limits^2_1 {f(x)} \, dx$ bằng 30/04/2023 cho f(x) liên tục trên R thỏa f(2x) – xf($x^{2}$) = $x^{3}$. khi đó $\int\limits^2_1 {f(x)} \, dx$ bằng
$\begin{array}{l}f\left( {2x} \right) – xf\left( {{x^2}} \right) = {x^3}\\ \Leftrightarrow 2f\left( {2x} \right) – 2xf\left( {{x^2}} \right) = 2{x^3}\\ \Leftrightarrow \int\limits_0^1 {2f\left( {2x} \right)dx} – \int\limits_0^1 {2xf\left( {{x^2}} \right)dx} = \int\limits_0^1 {2{x^3}dx} \\t = 2x \Rightarrow dt = 2dx,x = 0 \Rightarrow t = 0,x = 1 \Rightarrow t = 2\\u = {x^2} \Rightarrow du = 2xdx,x = 0 \Rightarrow u = 0,x = 1 \Rightarrow u = 1\\ \Leftrightarrow \int\limits_0^2 {f\left( t \right)dt – \int\limits_0^1 {f\left( u \right)du} = \left. {2.\frac{{{x^4}}}{4}} \right|} _0^1\\ \Leftrightarrow \int\limits_0^2 {f\left( x \right)dx} – \int\limits_0^1 {f\left( x \right)dx} = \frac{1}{2}\\ \Leftrightarrow \int\limits_1^2 {f\left( x \right)dx} = \frac{1}{2}\end{array}$ Trả lời
f\left( {2x} \right) – xf\left( {{x^2}} \right) = {x^3}\\
\Leftrightarrow 2f\left( {2x} \right) – 2xf\left( {{x^2}} \right) = 2{x^3}\\
\Leftrightarrow \int\limits_0^1 {2f\left( {2x} \right)dx} – \int\limits_0^1 {2xf\left( {{x^2}} \right)dx} = \int\limits_0^1 {2{x^3}dx} \\
t = 2x \Rightarrow dt = 2dx,x = 0 \Rightarrow t = 0,x = 1 \Rightarrow t = 2\\
u = {x^2} \Rightarrow du = 2xdx,x = 0 \Rightarrow u = 0,x = 1 \Rightarrow u = 1\\
\Leftrightarrow \int\limits_0^2 {f\left( t \right)dt – \int\limits_0^1 {f\left( u \right)du} = \left. {2.\frac{{{x^4}}}{4}} \right|} _0^1\\
\Leftrightarrow \int\limits_0^2 {f\left( x \right)dx} – \int\limits_0^1 {f\left( x \right)dx} = \frac{1}{2}\\
\Leftrightarrow \int\limits_1^2 {f\left( x \right)dx} = \frac{1}{2}
\end{array}$