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

The integral from 1 to 4 of (2x^2-x+1) is 75/2

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

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

\frac{75}{2}

Decimal

37.5

Solution steps

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

\int_{1}^{4} 2 x^{2} d x = 42

\int_{1}^{4} x d x = \frac{15}{2}

\int_{1}^{4} 1 d x = 3

= 42 - \frac{15}{2} + 3

Simplify

42 - \frac{15}{2} + 3 : \frac{75}{2}

= \frac{75}{2}

\int_{0}^{\frac{L}{2}} x^{2} d x

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

\int_{0}^{2} 12 x^{3} 9 + x^{4} d x

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

\int_{0}^{1} \frac{1}{x ^{0.9}} d x

What is the integral from 1 to 4 of (2x^2-x+1) ?
The integral from 1 to 4 of (2x^2-x+1) is 75/2