The graphs of functions f(x)=e^x and h(x)=e^(-.5x), where x is a real number and 0<x<1 ,lie on a plane. Draw these functions and find the area they and the line x=0.6 enclose using integration correct to 3 decimal places

∫ f(x)dx [.6, 0] - ∫ h(x)dx [.6,0]∫ f(x)dx = e^x....... f(x)dx [.6, 0] = (e^.6)-(e^0)=.822∫ h(x)dx = -2e^(-.5x)........... ∫ h(x)dx [.6,0]= (-2e^-.3)-(-2e^0)= .518.822 - .518= .304Therefore area enclosed = .518 units squared

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Answered by Brendan B. Maths tutor

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