What is the integral from 2 to y of 1/2 (r-2) ?

The integral from 2 to y of 1/2 (r-2) is (y^2-4y+4)/4

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integral from 2 to y of 1/2 (r-2)

\int_{2}^{y} \frac{1}{2} (r - 2) d r

\frac{y ^{2} - 4 y + 4}{4}

Solution steps

\int_{2}^{y} \frac{1}{2} (r - 2) d r

Take the constant out:

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

= \frac{1}{2} \cdot \int_{2}^{y} - 2 + r d r

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (- \int_{2}^{y} 2 d r + \int_{2}^{y} r d r)

\int_{2}^{y} 2 d r = 2 y - 4

\int_{2}^{y} r d r = \frac{y ^{2}}{2} - 2

= \frac{1}{2} (- (2 y - 4) + \frac{y ^{2}}{2} - 2)

Simplify

\frac{1}{2} (- (2 y - 4) + \frac{y ^{2}}{2} - 2) : \frac{y ^{2} - 4 y + 4}{4}

= \frac{y ^{2} - 4 y + 4}{4}

\int_{1}^{3} (3 x^{2} - 4 x + 1) d x

\int_{- 1}^{0} \frac{x}{e ^{x}} d x

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

\int_{- \infty}^{\infty} 2 d x

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

What is the integral from 2 to y of 1/2 (r-2) ?
The integral from 2 to y of 1/2 (r-2) is (y^2-4y+4)/4