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

The integral from 1 to 4 of (3x^3+x+1) is 807/4

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

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

\frac{807}{4}

Decimal

201.75

Solution steps

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

\int_{1}^{4} 3 x^{3} d x = \frac{765}{4}

\int_{1}^{4} x d x = \frac{15}{2}

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

= \frac{765}{4} + \frac{15}{2} + 3

Simplify

\frac{765}{4} + \frac{15}{2} + 3 : \frac{807}{4}

= \frac{807}{4}

What is the integral from 1 to 4 of (3x^3+x+1) ?
The integral from 1 to 4 of (3x^3+x+1) is 807/4