Cho hàm số f(x) liên tục trên [-1; 1] thỏa mãn f(x) - 1 = tích phân từ -1 đến 1

* Hướng dẫn giải

Ta có:

$f\left(x\right)-1=\underset{-1}{\overset{1}{\int }}\left(x+e^t\right)f\left(t\right)dt\iff f\left(x\right)-1=x\underset{-1}{\overset{1}{\int }}f\left(t\right)dt+\underset{-1}{\overset{1}{\int }}{e}^{t}f\left(t\right)dt\left(*\right)$

Giả sử $\underset{-1}{\overset{1}{\int }}f\left(t\right)dt=a,\underset{-1}{\overset{1}{\int }}{e}^{t}f\left(t\right)dt=b\Rightarrow f\left(x\right)-1=xa+b\iff f\left(x\right)=ax+b+1.$

Thay vào (*) ta có:

$ax+b=x\underset{-1}{\overset{1}{\int }}\left(at+b+1\right)dt+\underset{-1}{\overset{1}{\int }}{e}^t\left(at+b+1\right)dt$

$\iff ax+b=x\left(\frac{a{t}^2}{2}+bt+t\right)\left|\begin{array}{l}1\\ -1\end{array}\right.+\underset{-1}{\overset{1}{\int }}{e}^t\left(at+b+1\right)dt$

$\iff ax+b=x\left(\frac{a}{2}+b+1-\frac{a}{2}+b+1\right)+\left(at+b+1\right)e^t\left|{}_{-1}^1\right.-a\underset{-1}{\overset{1}{\int }}{e}^{t}dt$

$\iff ax+b=x\left(2b+2\right)+\left(a+b+1\right)e-\left(-a+b+1\right)e^{-1}-a\left(e-e^{-1}\right)$

$\iff ax+b=x\left(2b+2\right)+\left(b+1\right)e+\left(2a-b-1\right)e^{-1}$

$\Rightarrow \left\{\begin{array}{l}a=2b+2\\ b=\left(b+1\right)e+\left(2a-b-1\right)e^{-1}\end{array}\right.\iff \left\{\begin{array}{l}a=2b+2\\ b=\left(b+1\right)e+\left(3b+3\right)e^{-1}\end{array}\right.$

$\iff \left\{\begin{array}{l}a=2b+2\\ b=\left(e+\frac{3}{e}\right)b+e+\frac{3}{e}\end{array}\right.\iff \left\{\begin{array}{l}a=\frac{2e^2}{e^2-e-1}\\ b=\frac{e+\frac{3}{e}}{1-e-\frac{3}{e}}=\frac{e^2+3}{-e^2+e-3}\end{array}\right.$

Vậy $\underset{-1}{\overset{1}{\int }}{e}^{t}f\left(t\right)dt=\underset{-1}{\overset{1}{\int }}{e}^{x}f\left(x\right)dx=\frac{e^2+3}{-e^2+e-3}=I.$

Chọn C.