What is the integral of (b-|x|)^2 ?

The integral of (b-|x|)^2 is b^2x-bx|x|+(x^3)/3+C

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integral of (b-|x|)^2

\int (b - ∣ x ∣)^{2} d x

b^{2} x - b x ∣ x ∣ + \frac{x ^{3}}{3} + C

Solution steps

\int (b - ∣ x ∣)^{2} d x

Expand

(b - ∣ x ∣)^{2} : b^{2} - 2 b ∣ x ∣ + x^{2}

= \int b^{2} - 2 b ∣ x ∣ + 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 b^{2} d x - \int 2 b ∣ x ∣ d x + \int x^{2} d x

\int b^{2} d x = b^{2} x

\int 2 b ∣ x ∣ d x = b x ∣ x ∣

\int x^{2} d x = \frac{x ^{3}}{3}

= b^{2} x - b x ∣ x ∣ + \frac{x ^{3}}{3}

Add a constant to the solution

= b^{2} x - b x ∣ x ∣ + \frac{x ^{3}}{3} + C

f (x) = - 8 x^{2} - 7, at x = 1

\int ln (5 x + 8) d x

\frac{x ^{2} \cdot cos ( x )}{( 2 x + 1 ) ^{3}}

\int \frac{5 x + 2}{3 x 1 - 3 x} d x

x \to 0 lim (sin (π x))

What is the integral of (b-|x|)^2 ?
The integral of (b-|x|)^2 is b^2x-bx|x|+(x^3)/3+C