What is the integral from 0 to pi/2 of e^{it} ?

The integral from 0 to pi/2 of e^{it} is 1+i

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to pi/2 of e^{it}

\int_{0}^{\frac{π}{2}} e^{i t} d t

1 + i

Solution steps

\int_{0}^{\frac{π}{2}} e^{i t} d t

Apply u-substitution

: \int_{0}^{\frac{πi}{2}} - i e^{u} d u

= \int_{0}^{\frac{πi}{2}} - i e^{u} d u

Take the constant out:

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

= - i \cdot \int_{0}^{\frac{πi}{2}} e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= - i [e^{u}]_{0}^{\frac{πi}{2}}

Compute the boundaries:

i - 1

= - i (i - 1)

Simplify

= 1 + i

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

\int_{0}^{13} \frac{x}{16 + 5 x} d x

\int_{1}^{e^{1}} \frac{ln ( x )}{x} d x

\int_{- 11}^{0} 121 - x^{2} d x

\int_{0}^{4} 8 x - 2 x^{2} d x

What is the integral from 0 to pi/2 of e^{it} ?
The integral from 0 to pi/2 of e^{it} is 1+i