question 11 not yet answered marked out of 1.00 evaluate: $$ \\int _ { 0 } ^ { 2 } ( - x ^ { 2 } + 4 x ) - x…

Explanation:

Step1: Simplify the integrand

$$ \begin{align*} (-x^{2}+4x)-x^{2}&=-x^{2}+4x - x^{2}\ &=-2x^{2}+4x \end{align*} $$

Step2: Integrate term - by - term

Use the power rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$. For $\int(-2x^{2}+4x)dx=-2\int x^{2}dx + 4\int xdx$

• $-2\int x^{2}dx=-2\times\frac{x^{3}}{3}=-\frac{2}{3}x^{3}$
• $4\int xdx=4\times\frac{x^{2}}{2}=2x^{2}$

So, $\int(-2x^{2}+4x)dx=-\frac{2}{3}x^{3}+2x^{2}+C$

Step3: Evaluate the definite integral

Use the fundamental theorem of calculus $\int_{a}^{b}F^\prime(x)dx=F(b)-F(a)$, where $F(x)=-\frac{2}{3}x^{3}+2x^{2}$ $$ \begin{align*} \int_{0}^{2}(-2x^{2}+4x)dx&=\left(-\frac{2}{3}x^{3}+2x^{2}\right)\big|_{0}^{2}\ &=\left(-\frac{2}{3}(2)^{3}+2(2)^{2}\right)-\left(-\frac{2}{3}(0)^{3}+2(0)^{2}\right)\ &=\left(-\frac{16}{3}+8\right)-0\ &=\frac{-16 + 24}{3}\ &=\frac{8}{3} \end{align*} $$

Answer:

$\frac{8}{3}$