Chapter 6.3, Problem 51E

(a)

To determine

To calculate:the area of the region using the geometry formula.

Answer to Problem 51E

The $\text{area of region =}\frac{1}{2}bh$

Explanation of Solution

Given information:

Consider the region given below as,

  

Formula used:

  $\begin{array}{l}\text{area of region = area of traingle}\\ \text{=}\frac{1}{2}\left(b\right)\left(h\right)\end{array}$

Calculation:

Since this is a triangle.

  $\begin{array}{l}\text{area of region = area of traingle}\\ \text{=}\frac{1}{2}\left(b\right)\left(h\right)\\ \text{area of region =}\frac{1}{2}bh\end{array}$

Conclusion:

The $\text{area of region =}\frac{1}{2}bh$

(b)

To determine

To calculate:The antiderivative of y.

Answer to Problem 51E

The $\text{antiderivative of }y=\frac{h}{2b}{x}^2+C$

Explanation of Solution

Formula used:

  $\text{antiderivative of }y=\frac{h}{2b}{x}^2+C$

Calculation:

Generalized anti-derivative is given by

  $\text{antiderivative of }y=\frac{h}{2b}{x}^2+C$

Where C is a constant

Conclusion:

The $\text{antiderivative of }y=\frac{h}{2b}{x}^2+C$

(c)

To determine

To calculate:the anti derivative of yfor the given range

Answer to Problem 51E

The value of ${\displaystyle \underset{0}{\overset{b}{\int }}y\left(x\right)}dx=\frac{1}{2}hb$

Explanation of Solution

Formula used:

  ${\displaystyle \underset{a}{\overset{x}{\int }}f\left(t\right)}dt=F\left(x\right)-F\left(a\right)$

Calculation:

Take,

C=0

  $\text{antiderivative of }y=\frac{h}{2b}{x}^2$

Since,

  ${\displaystyle \underset{a}{\overset{x}{\int }}f\left(t\right)}dt=F\left(x\right)-F\left(a\right)$

Where F is any anti-derivative of f

So,

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{b}{\int }}y\left(x\right)}dx=\frac{h}{2b}{\left(b\right)}^2-\frac{h}{2b}{\left(0\right)}^2\\ =\frac{h{b}^2}{2b}-0\\ {\displaystyle \underset{0}{\overset{b}{\int }}y\left(x\right)}dx=\frac{1}{2}hb\end{array}$

Conclusion:

The value of ${\displaystyle \underset{0}{\overset{b}{\int }}y\left(x\right)}dx=\frac{1}{2}hb$