Lời giải:\(I = \int\limits_{ - \frac{\pi }{4}}^4 {f\left( x \right)dx} = \int\limits_{ - \frac{\pi }{4}}^0 {f\left( x \right)dx} + \int\limits_0^4 {f\left( x \right)dx} = \int\limits_{ - \frac{\pi }{4}}^0 {\left( {4a + {{\tan }^2}\,x} \right)dx} + \int\limits_0^4 {\left( {4x - \sqrt {4x + 9} } \right)dx} \)
\( = \int\limits_{ - \frac{\pi }{4}}^0 {\left( {4a - 1} \right)dx} + \int\limits_{ - \frac{\pi }{4}}^0 {\left( {1 + {{\tan }^2}\,x} \right)dx} + \int\limits_0^4 {\left( {4x - \sqrt {4x + 9} } \right)dx} \)
\( = \left. {\left( {4a - 1} \right)} \right|_{ - \frac{\pi }{4}}^0 + \left. {\tan \,x} \right|_{ - \frac{\pi }{4}}^0 + \left. {\left( {2{x^2} - \frac{{\sqrt {{{\left( {4x + 9} \right)}^3}} }}{6}} \right)} \right|_0^4\)
\( = \left( {4a - 1} \right)\pi + 1 + \frac{{47}}{3} = \frac{{50}}{3} \Leftrightarrow a = \frac{1}{4}\)
