What is the integral from 0 to 1 of 7cos(4x) ?

The integral from 0 to 1 of 7cos(4x) is 7/4 sin(4)

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integral from 0 to 1 of 7cos(4x)

\int_{0}^{1} 7 cos (4 x) d x

\frac{7}{4} sin (4)

Decimal Notation

- 1.32440 \dots

Solution steps

\int_{0}^{1} 7 cos (4 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 (4 x) d x

Apply u-substitution

: \int_{0}^{4} cos (u) \frac{1}{4} d u

= 7 \cdot \int_{0}^{4} cos (u) \frac{1}{4} d u

Take the constant out:

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

= 7 \cdot \frac{1}{4} \cdot \int_{0}^{4} cos (u) d u

Use the common integral:

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

= 7 \cdot \frac{1}{4} [sin (u)]_{0}^{4}

Simplify

7 \cdot \frac{1}{4} [sin (u)]_{0}^{4} : \frac{7}{4} [sin (u)]_{0}^{4}

= \frac{7}{4} [sin (u)]_{0}^{4}

Compute the boundaries:

sin (4)

= \frac{7}{4} sin (4)

\int \frac{2}{x} - \frac{3}{x ^{4}} d x

\frac{d}{d x} y + 3 y = 0

(5, - 3), (- 1, - 1)

x \to 0 lim (\frac{x - 45}{x - 5})

f (x) = \frac{x ^{2} - 3}{2}

What is the integral from 0 to 1 of 7cos(4x) ?
The integral from 0 to 1 of 7cos(4x) is 7/4 sin(4)