What is area f(x)=2sin(x),g(x)=2cos(x) ?

The answer to area f(x)=2sin(x),g(x)=2cos(x) is 4sqrt(2)

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area f(x)=2sin(x),g(x)=2cos(x)

area

f (x) = 2 sin (x), g (x) = 2 cos (x)

42

Solution steps

Apply the area formula:

\int_{\frac{π}{4}}^{\frac{5 π}{4}} ∣ 2 sin (x) - 2 cos (x) ∣ d x

Solve

\int_{\frac{π}{4}}^{\frac{5 π}{4}} ∣ 2 sin (x) - 2 cos (x) ∣ d x : 42

The area is:

= 42

y^{^{''}} + 2 y^{^{'}} + y = - 4 e^{- t}

y^{^{''}} + 9 y = 4 x e^{(2 x)}, y (0) = 0, y^{^{'}} (0) = 0

x \to 0 + lim (5 x + 9)

\int_{- 2}^{1} (x^{3} - 3 x + 2) d x

\frac{\partial}{\partial x} (\frac{x ^{2}}{y ^{2}})

What is area f(x)=2sin(x),g(x)=2cos(x) ?
The answer to area f(x)=2sin(x),g(x)=2cos(x) is 4sqrt(2)