What is area g(x)=cos(x)+1,f(x)=sin(x) ?

The answer to area g(x)=cos(x)+1,f(x)=sin(x) is 2+pi/2-pi

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area g(x)=cos(x)+1,f(x)=sin(x)

area

g (x) = cos (x) + 1, f (x) = sin (x)

2 + \frac{π}{2} - π

Decimal Notation

0.42920 \dots

Solution steps

Apply the area formula:

\int_{\frac{π}{2}}^{π} ∣ cos (x) + 1 - sin (x) ∣ d x

Solve

\int_{\frac{π}{2}}^{π} ∣ cos (x) + 1 - sin (x) ∣ d x : 2 + \frac{π}{2} - π

The area is:

= 2 + \frac{π}{2} - π

\int_{1}^{9} 3 - x d x

f (x) = x^{3} (x - 5)^{2}

x \to 9 - lim (\frac{∣ x - 9 ∣}{x - 9})

\int x^{11} ln (x) d x

What is area g(x)=cos(x)+1,f(x)=sin(x) ?
The answer to area g(x)=cos(x)+1,f(x)=sin(x) is 2+pi/2-pi