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

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

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

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

4

Solution steps

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

\int_{- 1}^{1} 4 (1 - x^{2}) d x = \frac{16}{3}

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

= \frac{16}{3} - \frac{4}{3}

Simplify

\frac{16}{3} - \frac{4}{3} : 4

= 4

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