# Math Help - graphs of antiderivatives!

1. ## graphs of antiderivatives!

can someone show me the antiderivative of this graph...thanks...

mathaction

2. Start drawing vertical lines.

At x = 0, G(x) = 0

As x increases from x = 0, you get trapeziods. One parallel length obviously is 10. The other is obtained from the equation of the line between (0,-10) and (10,-20). Calculate the area of ALL such trapezoids, depending on the value of x.

Of course, this piece will be negative.

Do the same with the next line segment, starting where the last piece left off.

3. Hello, mathaction!

Edit: I'll re-do this solution with the new information: g(0) = 50

Can someone show me the antiderivative of this graph?

For reference, label the points: . $A(0,\text{-}10),\;B(10,\text{-}20),\;C(15,0),\;D(20,10),\;E(40,0)$

Segment AB has slope: . $m \:=\:\frac{(\text{-}20) - (\text{-}10)}{10 - 0} \:=\:-1$ . and contains $A(0,\text{-}10)$

. . Its equation is: . $y -(\text{-}10) \:=\:\text{-}1(x-0)\quad\Rightarrow\quad y \:=\:\text{- }x-10$

We have: . $g'(x) \:=\:\text{- }x-10$

. . Integrate: . $g(x)\:=\;\text{- }\frac{1}{2}x^2-10x + C_1$

Since $g(0) = 50$, we have: . $50 \:=\:-\frac{1}{2}(0^2) - 10(0) + C_1\quad\Rightarrow\quad C_1 \:=\:50$

. . Therefore: ${\color{blue}\text{on }[0,10]\!:\;\;g(x) \:=\;\text{- }\frac{1}{2}x^2 - 10x + 50}$

Note that: . $g(10) \:=\:\text{- }\frac{1}{2}(10^2) - 10(10) + 50 \:=\:\text{- }100$

Segment BC has slope: . $m \:=\:\frac{0 - (\text{-}20)}{15-10} \:=\:4$ . and contains $C(15,0)$

. . Its equation is: . $y - 0 \:=\:4(x-15)\quad\Rightarrow\quad y \:=\:4x-60$

We have: . $g'(x) \:=\:4x-60$

. . Integrate: . $g(x)\:=\;2x^2-60x + C_2$

Since $g(10) = \text{- }100$, we have: . $\text{- }100 \:=\:2(10^2) - 60(10) + C_2\quad\Rightarrow\quad C_2 = 300$

. . Therefore: . ${\color{blue}\text{on }[10,\,20]\!:\;\;g(x) \:=\:2x^2 - 60x - 100}$

Note that: . $g(20) \:=\:2(20^2) - 60(20) + 300 \:=\:\text{- }100$

Segment CD has slope: . $m \:=\:\frac{10 - 0}{20 - 15} \:=\:2$ . and contains $C(15,0)$

. . Its equation is: . $y - 0 \:=\:21(x-15)\quad\Rightarrow\quad y \:=\:2x-30$

We have: . $g'(x) \:=\:2x-30$

. . Integrate: . $g(x)\:=\;x^2-30x + C_3$

Since $g(20) = \text{- }100\!:\;\text{- }100 \:=\:20^2 - 30(20) + C_3\quad\Rightarrow\quad C_3 = 100$

. . Therefore: . ${\color{blue}\text{on }[15,20],\;g(x) \;=\;x^2-30x + 100}$

Note that: . $g(20) \:=\:20^2 - 30(20) + 100 \:=\:\text{- }100$

Segment DE has slope: . $m \:=\:\frac{0-10}{40 - 20} \:=\:\text{- }\frac{1}{2}$ . and contains $E(40,0)$

. . Its equation is: . $y -0 \:=\:\text{- }\frac{1}{2}(x-40)\quad\Rightarrow\quad y \:=\:\text{- }\frac{1}{2}x+10$

We have: . $g'(x) \:=\:\text{- }\frac{1}{2}x+10$

. . Integrate: . $g(x)\:=\;\text{- }\frac{1}{4}x^2+10x + C_4$

Since $g(20) = \text{- }100$, we have: . $\text{- }100 \:=\:-\frac{1}{4}(20^2) + 10(20) + C_4\quad\Rightarrow\quad C_4 = \text{- }200$

. . Therefore: . ${\color{blue}\text{on }[20,\,40]\!:\;\;g(x) \:=\:\text{- }\frac{1}{4}x^2 + 10x - 200}$

We have a piecewise function: . $g(x) \;=\;\begin{Bmatrix}\text{- }\frac{1}{2}x^2 - 10x + 50 & 0 \leq x \leq 10 \\
2x^2-60x - 100 & 10 \leq x \leq 15 \\
x^2-30x + 100 & 15 \leq x \leq 20 \\
\text{- }\frac{1}{4}x^2+10x - 200 & 20 \leq x \leq 40
\end{Bmatrix}$

4. yeah sorry it does give g(0) it equals 50