What is the integral from 0 to 1 of (x^2+3x+4) ?

The integral from 0 to 1 of (x^2+3x+4) is 35/6

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integral from 0 to 1 of (x^2+3x+4)

\int_{0}^{1} (x^{2} + 3 x + 4) d x

\frac{35}{6}

Decimal Notation

5.83333 \dots

Solution steps

\int_{0}^{1} x^{2} + 3 x + 4 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_{0}^{1} x^{2} d x + \int_{0}^{1} 3 x d x + \int_{0}^{1} 4 d x

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

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

\int_{0}^{1} 4 d x = 4

= \frac{1}{3} + \frac{3}{2} + 4

Simplify

\frac{1}{3} + \frac{3}{2} + 4 : \frac{35}{6}

= \frac{35}{6}

x \to - 3 lim (\frac{3 - x}{3 + x})

2 (ln (x) + 1)

\int \frac{1}{x ^{e}} d x

y = 6 lo g_{5} (x^{2} - 6)

y = (5 - 2 x^{3})^{3}

What is the integral from 0 to 1 of (x^2+3x+4) ?
The integral from 0 to 1 of (x^2+3x+4) is 35/6