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

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

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

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

\frac{4}{3}

Decimal

1.33333 \dots

Solution steps

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

\int_{0}^{4} x^{2} d x = \frac{64}{3}

\int_{0}^{4} 3 x d x = 24

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

= \frac{64}{3} - 24 + 4

Simplify

\frac{64}{3} - 24 + 4 : \frac{4}{3}

= \frac{4}{3}

\int_{0}^{4} x^{3} d x

\int_{1}^{3} \frac{1}{( x - 1 ) ^{\frac{2}{3}}} d x

\int_{1}^{4} \frac{1}{( 3 - 5 x ) ^{2}} d x

\int_{0}^{l n (2)} e^{5 x} d x

\int_{0}^{6} x^{2} - 4 d x

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