What is the integral from 0 to 1 of 7cos(2x) ?

The integral from 0 to 1 of 7cos(2x) is 7/2 sin(2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 1 of 7cos(2x)

\int_{0}^{1} 7 cos (2 x) d x

\frac{7}{2} sin (2)

Solution steps

\int_{0}^{1} 7 cos (2 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 (2 x) d x

Apply u-substitution

: \int_{0}^{2} cos (u) \frac{1}{2} d u

= 7 \cdot \int_{0}^{2} cos (u) \frac{1}{2} d u

Take the constant out:

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

= 7 \cdot \frac{1}{2} \cdot \int_{0}^{2} cos (u) d u

Use the common integral:

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

= 7 \cdot \frac{1}{2} [sin (u)]_{0}^{2}

Simplify

7 \cdot \frac{1}{2} [sin (u)]_{0}^{2} : \frac{7}{2} [sin (u)]_{0}^{2}

= \frac{7}{2} [sin (u)]_{0}^{2}

Compute the boundaries:

sin (2)

= \frac{7}{2} sin (2)

\int 5 sin (5 θ) sin (θ) d θ

\int \frac{x}{x ^{2} + 5 x + 6} d x

\frac{d}{d x} (\frac{2}{π})

\int \frac{e ^{2 x}}{4 + e ^{4 x}} d x

\frac{d}{d x} (3 x^{2} - 18 x - 21)

What is the integral from 0 to 1 of 7cos(2x) ?
The integral from 0 to 1 of 7cos(2x) is 7/2 sin(2)