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

The integral from 1 to 4 of (x^2+sqrt(x)) is 77/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to 4 of (x^2+sqrt(x))

\int_{1}^{4} (x^{2} + x) d x

\frac{77}{3}

Decimal Notation

25.66666 \dots

Solution steps

\int_{1}^{4} x^{2} + x d x

Apply the Sum Rule:

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

= \int_{1}^{4} x^{2} d x + \int_{1}^{4} x d x

\int_{1}^{4} x^{2} d x = 21

\int_{1}^{4} x d x = \frac{14}{3}

= 21 + \frac{14}{3}

Simplify

21 + \frac{14}{3} : \frac{77}{3}

= \frac{77}{3}

\frac{\partial}{\partial x} ((x - 1)^{\frac{2}{3}} (y - 4))

\frac{d}{d y} (y ln (y))

f (x) = 6 e^{x} cos (x), (0, 6)

\frac{d ^{2}}{d x ^{2}} (3 csc (x))

\frac{\partial}{\partial x} (cos (x^{2} + x y))

What is the integral from 1 to 4 of (x^2+sqrt(x)) ?
The integral from 1 to 4 of (x^2+sqrt(x)) is 77/3