Question

Cho hàm số f(x) = 2x. Tính và so sánh kết quả: \(\int\limits^2_0f\left(x\right)dx\) và \(\int\limits^1_0f\left(x\right)dx+\int\limits^2_1f\left(x\right)dx\).

Answer 1

Ta có

\(\int\limits_0^2 {f\left( x \right)dx}  = \int\limits_0^2 {2xdx}  = 2\int\limits_0^2 {xdx}  = 2\left. {\left( {\frac{{{x^2}}}{2}} \right)} \right|_0^2 = 2\left( {\frac{{{2^2}}}{2} - \frac{{{0^2}}}{2}} \right) = 4\)

\(\int\limits_0^1 {f\left( x \right)dx}  + \int\limits_1^2 {f\left( x \right)dx}  = \int\limits_0^1 {2xdx}  + \int\limits_1^2 {2xdx}  = 2\left( {\int\limits_0^1 {xdx}  + \int\limits_1^2 {xdx} } \right) = 2\left[ {\left. {\left( {\frac{{{x^2}}}{2}} \right)} \right|_0^1 + \left. {\left( {\frac{{{x^2}}}{2}} \right)} \right|_1^2} \right]\)\( = 2\left[ {\left( {\frac{{{1^2}}}{2} - \frac{{{0^2}}}{2}} \right) + \left( {\frac{{{2^2}}}{2} - \frac{{{1^2}}}{2}} \right)} \right] = 4\)

Vậy \(\int\limits_0^2 {f\left( x \right)dx}  = \int\limits_0^1 {f\left( x \right)dx}  + \int\limits_1^2 {f\left( x \right)dx} \)