What is the integral from 0 to 1 of 7cos(3x) ?

The integral from 0 to 1 of 7cos(3x) is 7/3 sin(3)

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integral from 0 to 1 of 7cos(3x)

\int_{0}^{1} 7 cos (3 x) d x

\frac{7}{3} sin (3)

Decimal

0.32928 \dots

Solution steps

\int_{0}^{1} 7 cos (3 x) d x

Take the constant out:

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

= 7 \cdot \int_{0}^{1} cos (3 x) d x

Apply u-substitution

= 7 \cdot \int_{0}^{3} cos (u) \frac{1}{3} d u

Take the constant out:

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

= 7 \cdot \frac{1}{3} \cdot \int_{0}^{3} cos (u) d u

Use the common integral:

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

= 7 \cdot \frac{1}{3} [sin (u)]_{0}^{3}

Simplify

7 \cdot \frac{1}{3} [sin (u)]_{0}^{3} : \frac{7}{3} [sin (u)]_{0}^{3}

= \frac{7}{3} [sin (u)]_{0}^{3}

Compute the boundaries:

sin (3)

= \frac{7}{3} sin (3)

\frac{d}{d θ} (θ)

\frac{\partial}{\partial x} (sin (7 x^{2} y - 3 x y^{3}))

\frac{d}{d x} (cosh (x) cos (x^{2}))

\frac{d}{d x} (arctan (\frac{8}{x}))

\frac{d}{d x} (x + x + 1)

What is the integral from 0 to 1 of 7cos(3x) ?
The integral from 0 to 1 of 7cos(3x) is 7/3 sin(3)