EXERCISE 7.5

EXERCISE 7.5 9. $$ \begin{aligned} & \int \frac{d x}{x^{2}-3 x-4} \\ & \frac{1}{x^{2}-3 x-4}=\frac{1}{x^{2}-4 x+x-4}=\frac{1}{x(x-4)+1(x-4)} \\ & =\frac{1}{(x-4)(x+1)} \end{aligned} $$ Partial fraction decomposition $$ \frac{1}{(x-4)(x+1)}=\frac{A}{x-4}+\frac{B}{x+1} \quad \text { (1) } $$ Multiplying by $(x-4)(x+1)$ $$ 1=A(x+1)+B(x-4) \quad \text { (2) } $$ Put $x+1=0 \Rightarrow x=-1$ into (2) $$ \begin{aligned} & 1=A(-1+1)+B(-1-4) \\ & 1=-5 … Read more

EXERCISE 6.1

EXERCISE 6.1

  1-4. Find the area of the shaded region. (1)  The curve $y=f(x)=x^{2}+1$ lies above the curve $y=g(x)=x$ and the shaded region is bounded from left side at $x=-1$ and from the right side by the line $x=2$. $$ \begin{aligned} & A=\int_{a}^{b}[f(x)-g(x)] d x \\ \\ & A=\int_{-1}^{2}\left(x^{2}+1-x\right) d x \\ \\ & A=\left[\frac{x^{3}}{3}+x-\frac{x^{2}}{2}\right]_{-1}^{2} \\ … Read more