Integration with fraction

Question

I'm having difficulty understanding how below integral is evaluated :

$$\int_0^1\frac{(1-y)^2}{2}+(1-y)y dy = \frac{-1(1-y)^3}{6}+\frac{y^2}{2}-\frac{y^3}{3}\bigg|^1_0$$

What are steps involved in this evaluation ?

For $\frac{(1-y)^2}{2}$ this appears to be evaluated as $\frac{(1-y)^{2+1}}{(3)2}$ but I'm how unsure - 1 is inserted ?

Answer 1

By the chain rule,

$$((ax+b)^k)'=(ax+b)'k(ax+b)^{k-1}=ak(ax+b)^k$$

and from this,

$$\int(ax+b)^kdx=\frac{(ax+b)^{k+1}}{a(k+1)}.$$

Try with $a=-1$.

Answer 2

Antiderivative if $(1-y)^{2}$ is not $(1-y)^{3} /3$. If you differentiate the latter you get $-(1-y)^{2}$. Hence the antiderivative is $-(1-y)^{3} /3$.

Answer 3

Remember that where $f(x)$ is linear we have that:

$$\int[f(x)]^n dx=\frac{[f(x)]^{n+1}}{(n+1)f'(x)}+C$$

Answer 4

$$ \begin{align} I=&\int\bigg(\frac{(1-x)^2}{2}+x(1-x)\bigg)dx\\ =&\frac12\int(1-x)^2dx+\int x(1-x)dx\\ =&\frac12\int(1-x)^2dx+\int xdx-\int x^2dx\\ \end{align} $$ The first one: $$I_1=\int(1-x)^2dx$$ Substitution: $$u=1-x\Rightarrow -du=dx$$ Hence $$I_1=-\int u^2du$$ $$I_1=-\frac{u^3}3$$ $$I_1=-\frac{(1-x)^3}3$$ Second one: $$I_2=\int xdx=\frac{x^2}2$$ Third one: $$I_3=\int x^2dx=\frac{x^3}3$$ Put it all together: $$I=\frac12I_1+I_2-I_3$$ $$I=-\frac{(1-x)^3}6+\frac{x^2}2-\frac{x^3}3$$ Evaluating from $0$ to $1$ gives $$\int_0^1\bigg(\frac{(1-x)^2}{2}+x(1-x)\bigg)dx=\frac13$$