Cho hàm số \(y = f\left( x \right)\) có đồ thị trên \(\left[ { - 1;3} \right]\) như hình vẽ

Khi đó \(\int\limits_{ - 1}^3 {f\left( x \right)dx} \)

\(S = \int\limits_a^b {f\left( x \right)dx} \)   

\(S = \int\limits_0^b {f\left( x \right)dx} \)   

\(S = \int\limits_b^a {\left| {f\left( x \right)} \right|dx} \)

\(S = \int\limits_a^b {\left| {f\left( x \right)} \right|dx} \)

\(S = \int\limits_{ - 3}^{ - 1} {\left| {{x^2} - 1} \right|dx} \)

\(S = \int\limits_{ - 1}^{ - 3} {\left| {{x^2} - 1} \right|dx} \)

\(S = \int\limits_{ - 3}^0 {\left| {{x^2} - 1} \right|dx} \)

\(S = \int\limits_{ - 3}^{ - 1} {\left( {1 - {x^2}} \right)dx} \)

\(S = \int\limits_a^b {\left( {f\left( x \right) - g\left( x \right)} \right)dx} \)

\(S = \int\limits_a^b {\left( {g\left( x \right) - f\left( x \right)} \right)dx} \)

\(S = \int\limits_a^b {\left| {f\left( x \right) - g\left( x \right)} \right|dx} \)

\(S = \int\limits_a^b {\left| {f\left( x \right)} \right|dx}  - \int\limits_a^b {\left| {g\left( x \right)} \right|dx} \)

\(S = \int\limits_0^e {\left| {{e^x} + x} \right|dx} \)

\(S = \int\limits_0^e {\left| {{e^x} - x} \right|dx} \)

\(S = \int\limits_e^0 {\left| {{e^x} - x} \right|dx} \)

\(S = \int\limits_e^0 {\left| {{e^x} + x} \right|dx} \)

$S = \left| {\int_{ - 1}^1 {\left( {3x - {x^3}} \right)dx} } \right|$     

$S = \int_{ - 1}^0 {\left( {3x - {x^3}} \right)dx}  + \int_0^1 {\left( {{x^3} - 3x} \right)dx} $

$S = \int_{ - 1}^1 {\left( {3x - {x^3}} \right)dx} $    

$S = \int_{ - 1}^0 {\left( {{x^3} - 3x} \right)dx}  + \int_0^1 {\left( {3x - {x^3}} \right)dx} $

$S = 2 + e$     

$S = 2 - e$      

$S = e - 2$

$S = e - 1$

\(S = \dfrac{8}{3}\)

\(S = 1\)

$S=\dfrac{4}{3}$

\(S = 2\)