∫02(2ti+3t2j)dt=\int_0^2(2ti+3t^2j)dt=∫02(2ti+3t2j)dt= A2i+4jB4i+8jC4i-8jD2i-8jAnswer: B. 4i+8j Read Explanation: ∫02(2ti+3t2j)dt=[2t22i+3t33]02\int_0^2{(2ti+3t^2j)dt}= [\frac{2t^2}{2}i+\frac{3t^3}{3}]_0^2∫02(2ti+3t2j)dt=[22t2i+33t3]02=4i+8j=4i+8j=4i+8j Read more in App