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

The integral from 1 to 4 of x-1-(x^2-4x+3) is 9/2

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

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

\frac{9}{2}

Decimal

4.5

Solution steps

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

Simplify

= \int_{1}^{4} - x^{2} + 5 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_{1}^{4} x^{2} d x + \int_{1}^{4} 5 x d x - \int_{1}^{4} 4 d x

\int_{1}^{4} x^{2} d x = 21

\int_{1}^{4} 5 x d x = \frac{75}{2}

\int_{1}^{4} 4 d x = 12

= - 21 + \frac{75}{2} - 12

Simplify

- 21 + \frac{75}{2} - 12 : \frac{9}{2}

= \frac{9}{2}

\int_{0}^{3} 2 y^{2} - 6 y d y

\int_{1}^{3} π (\frac{1}{x})^{2} d x

\int_{0}^{\frac{π}{4}} 4 sec (θ) tan (θ) d θ

\int_{0}^{8} \frac{x}{9 + 5 x} d x

\int_{0}^{2} 1 - \frac{x ^{2}}{4} d x

What is the integral from 1 to 4 of x-1-(x^2-4x+3) ?
The integral from 1 to 4 of x-1-(x^2-4x+3) is 9/2