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

The integral from 1 to 4 of (3x^2+sqrt(x)) is 203/3

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integral from 1 to 4 of (3x^2+sqrt(x))

\int_{1}^{4} (3 x^{2} + x) d x

\frac{203}{3}

Decimal Notation

67.66666 \dots

Solution steps

\int_{1}^{4} 3 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} 3 x^{2} d x + \int_{1}^{4} x d x

\int_{1}^{4} 3 x^{2} d x = 63

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

= 63 + \frac{14}{3}

Simplify

63 + \frac{14}{3} : \frac{203}{3}

= \frac{203}{3}

\frac{\partial}{\partial x} (cos (x z) + sin (yz) - z)

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

\int \frac{5}{y} d y

\frac{d}{d x} (5 cos (3 x))

What is the integral from 1 to 4 of (3x^2+sqrt(x)) ?
The integral from 1 to 4 of (3x^2+sqrt(x)) is 203/3