What is the integral from 1 to 2 of tsqrt(t-1) ?

The integral from 1 to 2 of tsqrt(t-1) is 16/15

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to 2 of tsqrt(t-1)

\int_{1}^{2} t t - 1 d t

\frac{16}{15}

Decimal Notation

1.06666 \dots

Solution steps

\int_{1}^{2} t t - 1 d t

Apply u-substitution

: \int_{0}^{1} (u + 1) u d u

= \int_{0}^{1} (u + 1) u d u

Expand

(u + 1) u : u^{\frac{3}{2}} + u

= \int_{0}^{1} u^{\frac{3}{2}} + u d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{1} u^{\frac{3}{2}} d u + \int_{0}^{1} u d u

\int_{0}^{1} u^{\frac{3}{2}} d u = \frac{2}{5}

\int_{0}^{1} u d u = \frac{2}{3}

= \frac{2}{5} + \frac{2}{3}

Simplify

\frac{2}{5} + \frac{2}{3} : \frac{16}{15}

= \frac{16}{15}

x \to c lim (\frac{27}{18})

e^{6 x}

\frac{\partial}{\partial x} (\frac{( 3 x + 8 )}{4 y + 1})

\int_{0}^{1} (6 x + 1) d x

\int 3 x (cos (4 x)) d x

What is the integral from 1 to 2 of tsqrt(t-1) ?
The integral from 1 to 2 of tsqrt(t-1) is 16/15