What is the integral from-4 to 4 of (-x^2+16) ?

The integral from-4 to 4 of (-x^2+16) is 256/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from-4 to 4 of (-x^2+16)

\int_{- 4}^{4} (- x^{2} + 16) d x

\frac{256}{3}

Decimal

85.33333 \dots

Solution steps

\int_{- 4}^{4} - x^{2} + 16 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_{- 4}^{4} x^{2} d x + \int_{- 4}^{4} 16 d x

\int_{- 4}^{4} x^{2} d x = \frac{128}{3}

\int_{- 4}^{4} 16 d x = 128

= - \frac{128}{3} + 128

Simplify

- \frac{128}{3} + 128 : \frac{256}{3}

= \frac{256}{3}

\int_{1}^{5} (5 + \frac{1}{x} + e^{x}) d x

\int_{4.9}^{7.1} (12 x^{2} - 6 x + 4) d x

\int_{1}^{3} (2 x^{2} - 8) d x

\int_{- \infty}^{5} θ e^{θ} d θ

\int_{3}^{x} 9 - y^{2} d y

What is the integral from-4 to 4 of (-x^2+16) ?
The integral from-4 to 4 of (-x^2+16) is 256/3