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

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

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

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

\frac{3}{2}

Solution steps

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

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

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

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

= 1 - \frac{1}{2} + 1

Simplify

1 - \frac{1}{2} + 1 : \frac{3}{2}

= \frac{3}{2}

\int_{2}^{5} 2000 x + 1000 d x

\int_{- 2}^{5} x^{2} - 3 x - 10 d x

\int_{5}^{10} \frac{2 t}{t ^{2} + 1} d t

\int_{- 2}^{1} \frac{t}{( t ^{2} + 1 ) ^{2}} d t

\int_{0}^{2 π} x^{4} + 4 x^{2} d x

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