What is the integral from 0 to 1 of (7x^2-4x+2) ?

The integral from 0 to 1 of (7x^2-4x+2) is 7/3

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integral from 0 to 1 of (7x^2-4x+2)

\int_{0}^{1} (7 x^{2} - 4 x + 2) d x

\frac{7}{3}

Decimal

2.33333 \dots

Solution steps

\int_{0}^{1} 7 x^{2} - 4 x + 2 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} 7 x^{2} d x - \int_{0}^{1} 4 x d x + \int_{0}^{1} 2 d x

\int_{0}^{1} 7 x^{2} d x = \frac{7}{3}

\int_{0}^{1} 4 x d x = 2

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

= \frac{7}{3} - 2 + 2

Simplify

= \frac{7}{3}

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

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

\int_{0}^{\frac{π}{4}} (2 cos^{2} (x)) d x

\int_{- π}^{0} x cos ((2 n + 1) x) d x

\int_{7}^{2} x d x

What is the integral from 0 to 1 of (7x^2-4x+2) ?
The integral from 0 to 1 of (7x^2-4x+2) is 7/3