question the shaded region shown below is bounded by the functions f(x)=−4x² + 4 and g(x)=1.75x + 2, the y…

Explanation:

Step1: Determine the upper - lower functions

On the interval $[0,1]$, we need to find which function is on top. Let's find the difference $h(x)=f(x)-g(x)=-4x^{2}+4-(1.75x + 2)=-4x^{2}-1.75x + 2$. We can also test a value in the interval $[0,1]$, say $x = 0$. $f(0)=4$ and $g(0)=2$, so $f(x)\geq g(x)$ on $[0,1]$.

Step2: Use the definite - integral formula for area

The area $A$ between two curves $y = f(x)$ and $y = g(x)$ from $x=a$ to $x = b$ is given by $A=\int_{a}^{b}[f(x)-g(x)]dx$. Here, $a = 0$, $b = 1$, $f(x)=-4x^{2}+4$ and $g(x)=1.75x + 2$. So $A=\int_{0}^{1}((-4x^{2}+4)-(1.75x + 2))dx=\int_{0}^{1}(-4x^{2}-1.75x + 2)dx$.

Step3: Integrate term - by - term

We know that $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$). $\int_{0}^{1}(-4x^{2}-1.75x + 2)dx=\left[-4\times\frac{x^{3}}{3}-1.75\times\frac{x^{2}}{2}+2x\right]_{0}^{1}$.

Step4: Evaluate the definite integral

Substitute the upper and lower limits: $\left(-\frac{4}{3}\times1^{3}-\frac{1.75}{2}\times1^{2}+2\times1\right)-\left(-\frac{4}{3}\times0^{3}-\frac{1.75}{2}\times0^{2}+2\times0\right)$ $=-\frac{4}{3}-\frac{1.75}{2}+2$ $=-\frac{4}{3}-\frac{7}{8}+2$ $=\frac{-32 - 21+48}{24}=\frac{-53 + 48}{24}=\frac{-5}{24}\approx - 0.208$ (This is wrong. We should have $g(x)-f(x)$ since $g(x)\geq f(x)$ on part of $[0,1]$. Let's start over.)

The correct way: We need to find the intersection of $y=-4x^{2}+4$ and $y = 1.75x + 2$. Set $-4x^{2}+4=1.75x + 2$, or $4x^{2}+1.75x - 2=0$. Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ with $a = 4$, $b=1.75$, $c=-2$. $x=\frac{-1.75\pm\sqrt{(1.75)^{2}-4\times4\times(-2)}}{2\times4}=\frac{-1.75\pm\sqrt{3.0625 + 32}}{8}=\frac{-1.75\pm\sqrt{35.0625}}{8}=\frac{-1.75\pm5.921}{8}$. The positive root is $x=\frac{-1.75 + 5.921}{8}\approx0.521$.

On $[0,0.521]$, $f(x)\geq g(x)$ and on $[0.521,1]$, $g(x)\geq f(x)$.

$A=\int_{0}^{0.521}((-4x^{2}+4)-(1.75x + 2))dx+\int_{0.521}^{1}((1.75x + 2)-(-4x^{2}+4))dx$

$\int_{0}^{0.521}(-4x^{2}-1.75x + 2)dx=\left[- \frac{4x^{3}}{3}-\frac{1.75x^{2}}{2}+2x\right]_{0}^{0.521}$ $=-\frac{4}{3}(0.521)^{3}-\frac{1.75}{2}(0.521)^{2}+2(0.521)$

$\int_{0.521}^{1}(4x^{2}+1.75x - 2)dx=\left[\frac{4x^{3}}{3}+\frac{1.75x^{2}}{2}-2x\right]_{0.521}^{1}$

Using a calculator: $\int_{0}^{0.521}(-4x^{2}-1.75x + 2)dx\approx1.007$ $\int_{0.521}^{1}(4x^{2}+1.75x - 2)dx\approx0.208$

$A\approx1.007+0.208 = 1.215$

Answer:

$1.215$