What is the integral from 0 to 1 of 5cos((pit)/2) ?

The integral from 0 to 1 of 5cos((pit)/2) is (10)/pi

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integral from 0 to 1 of 5cos((pit)/2)

\int_{0}^{1} 5 cos (\frac{π t}{2}) d t

\frac{10}{π}

Decimal

3.18309 \dots

Solution steps

\int_{0}^{1} 5 cos (\frac{π t}{2}) d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 5 \cdot \int_{0}^{1} cos (\frac{π t}{2}) d t

Apply u-substitution

: \int_{0}^{\frac{π}{2}} cos (u) \frac{2}{π} d u

= 5 \cdot \int_{0}^{\frac{π}{2}} cos (u) \frac{2}{π} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 5 \cdot \frac{2}{π} \cdot \int_{0}^{\frac{π}{2}} cos (u) d u

Use the common integral:

\int cos (u) d u = sin (u)

= 5 \cdot \frac{2}{π} [sin (u)]_{0}^{\frac{π}{2}}

Simplify

5 \cdot \frac{2}{π} [sin (u)]_{0}^{\frac{π}{2}} : \frac{10}{π} [sin (u)]_{0}^{\frac{π}{2}}

= \frac{10}{π} [sin (u)]_{0}^{\frac{π}{2}}

Compute the boundaries:

1

= \frac{10}{π} \cdot 1

Simplify

= \frac{10}{π}

y^{^{''}} - 4 y^{^{'}} = 16 t

\int \frac{1}{( a + b x ) ^{\frac{1}{3}}} d x

x^{2} y d x + x^{3} d y = 0

x \to - 2 lim (- 2 x^{3} - 3 x^{2} + 8)

\int 0.1 e^{0.4 x} d x

What is the integral from 0 to 1 of 5cos((pit)/2) ?
The integral from 0 to 1 of 5cos((pit)/2) is (10)/pi