What is the integral from 0 to 1 of sqrt(1+2t^2+t^4) ?

The integral from 0 to 1 of sqrt(1+2t^2+t^4) is 4/3

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 sqrt(1+2t^2+t^4)

\int_{0}^{1} 1 + 2 t^{2} + t^{4} d t

\frac{4}{3}

Decimal Notation

1.33333 \dots

Solution steps

\int_{0}^{1} 1 + 2 t^{2} + t^{4} d t

Factor

1 + 2 t^{2} + t^{4} : (t^{2} + 1)^{2}

= \int_{0}^{1} (t^{2} + 1)^{2} d t

Apply exponent rule:

a^{2} = ∣ a ∣

= \int_{0}^{1} (t^{2} + 1)^{2} d t

Eliminate Absolutes

= \int_{0}^{1} (t^{2} + 1)^{2} d t

Apply exponent rule:

a^{2} = ∣ a ∣

= \int_{0}^{1} (t^{2} + 1) d t

Eliminate Absolutes

= \int_{0}^{1} t^{2} + 1 d t

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} t^{2} d t + \int_{0}^{1} 1 d t

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

\int_{0}^{1} 1 d t = 1

= \frac{1}{3} + 1

Simplify

\frac{1}{3} + 1 : \frac{4}{3}

= \frac{4}{3}

\int_{2}^{\infty} x^{- 4} d x

\int_{0}^{2} \frac{3}{8} (4 x - 2 x^{2}) d x

\int_{e}^{e^{2}} (ln (x))^{2} (\frac{1}{x}) d x

\int_{- 1}^{1} \frac{x + 1}{x - 1} d x

\int_{0}^{8} ∣ 4 - x ∣ d x

What is the integral from 0 to 1 of sqrt(1+2t^2+t^4) ?
The integral from 0 to 1 of sqrt(1+2t^2+t^4) is 4/3