What is the integral from y-2 to 2 of (2)(-(y-x)+2) ?

The integral from y-2 to 2 of (2)(-(y-x)+2) is y^2-8y+16

Popular >

integral from y-2 to 2 of (2)(-(y-x)+2)

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

y^{2} - 8 y + 16

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{- 2 + y}^{2} - (y - 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

= 2 (- \int_{- 2 + y}^{2} y - x d x + \int_{- 2 + y}^{2} 2 d x)

\int_{- 2 + y}^{2} y - x d x = 4 y - y^{2} - 2 + \frac{( - 2 + y ) ^{2}}{2}

\int_{- 2 + y}^{2} 2 d x = - 2 y + 8

= 2 (- (4 y - y^{2} - 2 + \frac{( - 2 + y ) ^{2}}{2}) - 2 y + 8)

Simplify

2 (- (4 y - y^{2} - 2 + \frac{( - 2 + y ) ^{2}}{2}) - 2 y + 8) : y^{2} - 8 y + 16

= y^{2} - 8 y + 16

\int_{- 1}^{1} \frac{x ^{2}}{x + 2} d x

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

\int_{- \infty}^{- 1} \frac{1}{x ^{5}} d x

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

\int_{0}^{\infty} \frac{1}{y ^{3}} d y

What is the integral from y-2 to 2 of (2)(-(y-x)+2) ?
The integral from y-2 to 2 of (2)(-(y-x)+2) is y^2-8y+16