Find the definite integral ∫40x√1+2xdx
Let u=1+2x, then du=−2dx, so dx=du2. When x=0,u=1 and when x=4,u=9. Thus,
∫40x√1+2xdx=∫40u−12√udu2∫40x√1+2xdx=∫40u−12√u⋅du2∫40x√1+2xdx=14∫40u−1√udu∫40x√1+2xdx=14∫40u√u−1√udu∫40x√1+2xdx=14∫40u12−u−12du∫40x√1+2xdx=14[u12+112+1−u−12+1−12+1]40∫40x√1+2xdx=14[u3232−u−1212]40∫40x√1+2xdx=14[2(9)323−2(9)12]−14[2(1)323−2(1)12]∫40x√1+2xdx=14(12)−14(−43)∫40x√1+2xdx=3+412∫40x√1+2xdx=3+13∫40x√1+2xdx=9+13∫40x√1+2xdx=103
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