What is area 5sin(x),5cos(2x),0, pi/2 ?

The answer to area 5sin(x),5cos(2x),0, pi/2 is (15sqrt(3))/2-5

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area 5sin(x),5cos(2x),0, pi/2

area

5 sin (x), 5 cos (2 x), 0, \frac{π}{2}

\frac{15 3}{2} - 5

Decimal Notation

7.99038 \dots

Solution steps

Apply the area formula:

\int_{0}^{\frac{π}{2}} ∣ 5 sin (x) - 5 cos (2 x) ∣ d x

Solve

\int_{0}^{\frac{π}{2}} ∣ 5 sin (x) - 5 cos (2 x) ∣ d x : \frac{15 3}{2} - 5

The area is:

= \frac{15 3}{2} - 5

\int_{- 3}^{3} \frac{1}{2} (81 - x^{4}) d x

\frac{d}{d x} (25 x^{2} - 1)

x \to \frac{π}{4} lim (tan^{t a n (2 x)} (x))

f (x) = sin (5 ln (x))

\frac{d v}{d v}

What is area 5sin(x),5cos(2x),0, pi/2 ?
The answer to area 5sin(x),5cos(2x),0, pi/2 is (15sqrt(3))/2-5