What is the integral from 1 to 0 of (3x^2+x-5) ?

The integral from 1 to 0 of (3x^2+x-5) is 7/2

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integral from 1 to 0 of (3x^2+x-5)

\int_{1}^{0} (3 x^{2} + x - 5) d x

\frac{7}{2}

Decimal Notation

3.5

Solution steps

\int_{1}^{0} 3 x^{2} + x - 5 d x

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{1} 3 x^{2} + x - 5 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} 5 d x)

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

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

\int_{0}^{1} 5 d x = 5

= - (1 + \frac{1}{2} - 5)

Simplify

- (1 + \frac{1}{2} - 5) : \frac{7}{2}

= \frac{7}{2}

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What is the integral from 1 to 0 of (3x^2+x-5) ?
The integral from 1 to 0 of (3x^2+x-5) is 7/2