Question

1. properties of integrals use only the fact that

int _ { 0 } ^ { 4 } 3 x ( 4 - x ) d x = 32 , and the definitions and properties of inte-
grals, to evaluate the following integrals, if possible.
( mathbf { a } ). ( int _ { 4 } ^ { 0 } 3 x ( 4 - x ) d x )
( mathbf { b } ). ( int _ { 0 } ^ { 4 } x ( x - 4 ) d x )
( mathbf { c } ). ( int _ { 4 } ^ { 0 } 6 x ( 4 - x ) d x )
( mathbf { d } ). ( int _ { 0 } ^ { 8 } 3 x ( 4 - x ) d x )

Explanation:

Part (a)

Step1: Recall the property of definite integrals: $\int_{a}^{b} f(x) dx = -\int_{b}^{a} f(x) dx$

We know that $\int_{0}^{4} 3x(4 - x) dx = 32$. For $\int_{4}^{0} 3x(4 - x) dx$, we apply the above property.
$\int_{4}^{0} 3x(4 - x) dx = -\int_{0}^{4} 3x(4 - x) dx$

Step2: Substitute the known value

Substitute $\int_{0}^{4} 3x(4 - x) dx = 32$ into the equation.
$\int_{4}^{0} 3x(4 - x) dx = - 32$

Step1: Rewrite the integrand

First, rewrite $x(x - 4)$ as $-x(4 - x)$. So the integral $\int_{0}^{4} x(x - 4) dx$ becomes $\int_{0}^{4} -x(4 - x) dx$

Step2: Use the constant multiple rule of integrals: $\int_{a}^{b} kf(x) dx = k\int_{a}^{b} f(x) dx$ (where $k$ is a constant)

Here, $k = -\frac{1}{3}$? Wait, no. Wait, the original integrand in the known integral is $3x(4 - x)$. Our current integrand is $-x(4 - x)=\frac{-1}{3}\times3x(4 - x)$
So $\int_{0}^{4} x(x - 4) dx=\int_{0}^{4}-x(4 - x) dx=\frac{-1}{3}\int_{0}^{4}3x(4 - x) dx$

Step3: Substitute the known value

We know that $\int_{0}^{4} 3x(4 - x) dx = 32$. Substitute this value in:
$\frac{-1}{3}\times32 = -\frac{32}{3}$

Step1: Use the property of definite integrals: $\int_{a}^{b} f(x) dx = -\int_{b}^{a} f(x) dx$ and constant multiple rule

First, $\int_{4}^{0} 6x(4 - x) dx = -\int_{0}^{4} 6x(4 - x) dx$ (by the property of reversing limits). Then, using the constant multiple rule $\int_{a}^{b} kf(x) dx = k\int_{a}^{b} f(x) dx$, where $k = 6$ and $f(x)=x(4 - x)$, and also note that $6x(4 - x)=2\times3x(4 - x)$
So $\int_{4}^{0} 6x(4 - x) dx = -\int_{0}^{4} 6x(4 - x) dx=- 2\int_{0}^{4}3x(4 - x) dx$

Step2: Substitute the known value

We know that $\int_{0}^{4} 3x(4 - x) dx = 32$. Substitute this value:
$-2\times32=-64$

Answer:

$\boldsymbol{-32}$

Part (b)