∫01(3f(x)+2g(x)) dx\displaystyle\int_0^1 (3 f(x) + 2 g(x))\,dx∫01(3f(x)+2g(x))dx if ∫01f=4\int_0^1 f = 4∫01f=4 and ∫01g=−1\int_0^1 g = -1∫01g=−1 equalsA101010check_circleB141414C333D111111Explanation3(4)+2(−1)=12−2=103(4) + 2(-1) = 12 - 2 = 103(4)+2(−1)=12−2=10.