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

The integral from 1 to 4 of (4x^2+sqrt(x)) is 266/3

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

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

\frac{266}{3}

Decimal Notation

88.66666 \dots

Solution steps

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

\int_{1}^{4} 4 x^{2} d x = 84

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

= 84 + \frac{14}{3}

Simplify

84 + \frac{14}{3} : \frac{266}{3}

= \frac{266}{3}

\int \frac{1}{1 - 5 t} d t

\frac{\partial}{\partial x} (x^{3} \cdot 4 x + 5 y)

y = x (x - 4)

\frac{d}{d x} (sin (x) + cos (2 x))

\int_{0}^{2} ∣ x - 1 ∣ d x

What is the integral from 1 to 4 of (4x^2+sqrt(x)) ?
The integral from 1 to 4 of (4x^2+sqrt(x)) is 266/3