What is the integral from-1 to 1 of 7x^2-4x-3 ?

The integral from-1 to 1 of 7x^2-4x-3 is -4/3

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integral from-1 to 1 of 7x^2-4x-3

\int_{- 1}^{1} 7 x^{2} - 4 x - 3 d x

- \frac{4}{3}

Decimal Notation

- 1.33333 \dots

Solution steps

\int_{- 1}^{1} 7 x^{2} - 4 x - 3 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}^{1} 7 x^{2} d x - \int_{- 1}^{1} 4 x d x - \int_{- 1}^{1} 3 d x

\int_{- 1}^{1} 7 x^{2} d x = \frac{14}{3}

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

\int_{- 1}^{1} 3 d x = 6

= \frac{14}{3} - 0 - 6

Simplify

\frac{14}{3} - 0 - 6 : - \frac{4}{3}

= - \frac{4}{3}

\int \frac{csc ^{2} ( x )}{3 - cot ( x )} d x

\frac{\partial}{\partial x} (- s^{2} t^{2})

t \frac{d y}{d t} + 5 y = t^{3}

\int \frac{1}{8 + 2 x - x ^{2}} d x

x \to \infty lim (8 x^{2})

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