Cho hai hàm số \(f\left( x \right) = {e^x}\) và \(g\left( x \right) = 2{e^x} - 3\).

a) \(\int\limits_0^{\ln 2} {g\left( x \right){\rm{d}}x = 2 - 3\ln 2} \).

b) \(2\int\limits_0^2 {f\left( x \right){\rm{d}}x}  = 3 + \int\limits_0^2 {g\left( x \right){\rm{d}}x} \).

c) \(\int\limits_2^7 {\left[ {2f\left( x \right) - g\left( x \right)} \right]} \,{\rm{d}}x =  - 15\).

d) Nếu \(\int\limits_0^1 {f\left( x \right)g\left( x \right){\rm{d}}x}  = a \cdot {e^2} + b \cdot e + c\) (với \(a,b,c\) là các số nguyên) thì \(a + b + c = 0\).

\[\int\limits_0^{\ln 2} {g\left( x \right){\rm{d}}x = \int\limits_0^{\ln 2} {\left( {2{e^x} - 3} \right){\rm{d}}x} = \left. {\left( {2{e^x} - 3x} \right)} \right|_0^{\ln 2} = 2 - 3\ln 2} \].

\(\int\limits_0^2 {g\left( x \right){\rm{d}}x} = \int\limits_0^2 {\left[ {2f\left( x \right) - 3} \right]{\rm{d}}x} = 2\int\limits_0^2 {f\left( x \right){\rm{d}}x} - 3\int\limits_0^2 {{\rm{d}}x} = 2\int\limits_0^2 {f\left( x \right){\rm{d}}x} - 6\).

Suy ra \(2\int\limits_0^2 {f\left( x \right){\rm{d}}x} = 6 + \int\limits_0^2 {g\left( x \right){\rm{d}}x} \).

\(\int\limits_2^7 {\left[ {2f\left( x \right) - g\left( x \right)} \right]} \,{\rm{d}}x = \int\limits_2^7 {\left[ {2{e^x} - \left( {2{e^x} - 3} \right)} \right]\,} {\rm{d}}x = \int\limits_2^7 {3{\rm{d}}x} = \left. {3x} \right|_2^7 = 15\).

\(\int\limits_0^1 {f\left( x \right)g\left( x \right){\rm{d}}x} = \int\limits_0^1 {\left( {2{e^{2x}} - 3{e^x}} \right){\rm{d}}x} = \left. {\left( {2\frac{{{e^{2x}}}}{{\ln {e^2}}} - 3{e^x}} \right)} \right|_0^1 = \left. {\left( {{e^{2x}} - 3{e^x}} \right)} \right|_0^1 = {e^2} - 3e + 2\).

Suy ra \(a = 1;b = - 3;c = 2\). Vậy \(a + b + c = 1 - 3 + 2 = 0\).

Đáp án:       a) Đúng,      b) Sai,                   c) Sai,                   d) Đúng.