Cho ∫02f(x)dx=3{\int_{0}^{2}f}(x)dx = 3∫02f(x)dx=3. Tính I=∫02(1+2f(x))dxI = {\int_{0}^{2}{(1 + 2f(}}x))dxI=∫02(1+2f(x))dx.
I = 7.
I = 4.
I = 8.
I = 6.
I=∫02(1+2f(x))dxI = {\int_{0}^{2}{(1 + 2f(x))dx}}I=∫02(1+2f(x))dx
=∫02dx+2∫02f(x)dx= {\int_{0}^{2}{dx}} + 2{\int_{0}^{2}{f(x)dx}}=∫02dx+2∫02f(x)dx
=2+2.3=8= 2 + 2.3 = 8=2+2.3=8.
