1. Problem statement: The graph shows $f(x)=0$ for $0\le x\le 1$ and a line from $(1,0)$ to $(10,3)$, and you are given $\int_4^{10} f(x)\,dx = 12$. Find the integrals (a) through (e). 2. Formula and rules: The definite integral equals the signed area under the curve. Triangle area formula: $A=\frac{1}{2}bh$. Trapezoid area formula: $A=\frac{b_1+b_2}{2}h$. Important rule: when simplifying fractions I will show cancellations using $\cancel{\;\;}$. 3. Determine $f(x)$ from the geometry: $$f(x)=\begin{cases}0 & 0\le x\le 1 \\ \frac{x-1}{3} & 1\le x\le 10\end{cases}$$ Compute values needed: $$f(1)=0$$ $$f(3)=\frac{3-1}{3}=\frac{2}{3}$$ $$f(4)=\frac{4-1}{3}=1$$ 4. (a) Compute $\displaystyle\int_0^3 f(x)\,dx$. The area from 0 to1 is zero and from 1 to3 is a triangle with base $2$ and height $\frac{2}{3}$. $$\text{Area}=\frac{1}{2}\cdot 2\cdot \frac{2}{3}$$ Show cancellation: $$\frac{1}{2}\cdot 2\cdot \frac{2}{3}=\frac{\cancel{2}}{\cancel{2}}\cdot \frac{2}{3}=\frac{2}{3}$$ So (a) $=\frac{2}{3}$. 5. (b) Compute $\displaystyle\int_3^4 f(x)\,dx$. This is a trapezoid (base $1$) with heights $f(3)=\frac{2}{3}$ and $f(4)=1$. $$\text{Area}=\frac{\frac{2}{3}+1}{2}\cdot 1=\frac{\frac{5}{3}}{2}=\frac{5}{6}$$ So (b) $=\frac{5}{6}$. 6. (c) Compute $\displaystyle\int_0^4 f(x)\,dx=\int_0^3 f+\int_3^4 f$. $$\frac{2}{3}+\frac{5}{6}=\frac{4}{6}+\frac{5}{6}=\frac{9}{6}$$ Show cancellation: $$\frac{9}{6}=\frac{\cancel{3}\cdot 3}{\cancel{3}\cdot 2}=\frac{3}{2}$$ So (c) $=\frac{3}{2}$. 7. (d) Given directly from the problem: $\displaystyle\int_4^{10} f(x)\,dx=12$. 8. (e) Compute $\displaystyle\int_0^{10}6f(x)\,dx=6\int_0^{10}f(x)\,dx=6\big(\int_0^4 f+\int_4^{10} f\big)=6\big(\tfrac{3}{2}+12\big)$. Combine inside: $$\frac{3}{2}+12=\frac{3}{2}+\frac{24}{2}=\frac{27}{2}$$ Show cancellation when multiplying: $$6\cdot\frac{27}{2}=\frac{6\cdot 27}{2}=\frac{\cancel{2}\cdot 3\cdot 27}{\cancel{2}}=3\cdot 27=81$$ So (e) $=81$. Final answers: (a) $\frac{2}{3}$ (b) $\frac{5}{6}$ (c) $\frac{3}{2}$ (d) $12$ (e) $81$