SL 5.5—Integration introduction, areas between curve and x axis

We use the power rule for integration: $\int a{x}^{n}dx=\frac{a{x}^{n+1}}{n+1}+C$. We integrate each term separately.

$\int 4x^{3}dx=\frac{4x^4}{4}=x^4$

$\int -6x^{2}dx=\frac{-6x^3}{3}=-2x^3,\int 5dx=5x$

Adding all terms and the constant of integration: $\int (4x^3-6x^2+5)dx=x^4-2x^3+5x+C$

We need the correct anti-derivative of $4x^3-6x^2+5$. The key checks are: correct power-rule application and inclusion of $+C$.

The option $12x^2-12x+C$ is actually the derivative of $4x^3-6x^2+5$, not the integral. Eliminate this.

The option $x^4-2x^3+5+C$ fails to integrate the constant term $5$ correctly — it should become $5x$, not $5$. Eliminate this.

The option $4x^4-6x^3+5x+C$ simply raised the powers without dividing by the new power. Eliminate this.

The remaining option, $x^4-2x^3+5x+C$, correctly applies 'add one to the power, divide by the new power' for each term.

The general anti-derivative is $F(x)=2x^3-2x^2+x+C$.

The curve passes through $(0,3)$, so $F(0)=3$: $3=2(0{)}^3-2(0{)}^2+(0)+C=C$

Since $C=3$, the specific anti-derivative is $F(x)=2x^3-2x^2+x+3$

We need to find $C$ by substituting $x=0$, $F(0)=3$ into $F(x)=2x^3-2x^2+x+C$. At $x=0$, all polynomial terms vanish, leaving $C=3$.

The option $2x^3-2x^2+x-3$ has $C=-3$, which would give $F(0)=-3\ne 3$. Eliminate.

The option $2x^3-2x^2+x$ has no constant term, meaning $F(0)=0\ne 3$. Eliminate.

The option $2x^3-2x^2+x+1$ gives $F(0)=1\ne 3$. Eliminate.

Only $2x^3-2x^2+x+3$ satisfies $F(0)=3$.