#### Passage

Let g(x) =$\displaystyle \int_{0}^{x}f\left ( t \right )dt,$ where f is a function
whose graph is show adjacently.
On the basis of above information, answer te following questions.
Mathematics

# Set of values of x in [0,7] for which g(x) is negative is

(5,7)

##### SOLUTION
$g(x)$ starts decreasing from $x=3$
$\displaystyle g\left ( 4 \right )=\int_{0}^{4}f\left ( t \right )dt=\int_{0}^{3}f\left ( t \right )dt+\int_{3}^{4}f\left ( t \right )dt$
$\displaystyle =\frac{9}{2}+\int_{3}^{4}\left ( -3t+9 \right )dt=\frac{9}{2}+\left ( 9t+\frac{3t^{2}}{2} \right )^{4}_{3}$
$\displaystyle \frac{9}{2}+\left ( 36-24-27+\frac{27}{2} \right )=3$ Now $\displaystyle g\left ( x \right )=\int_{0}^{x}f\left ( t \right )dt$
$\displaystyle =\int_{0}^{4}f\left ( t \right )dt+\int_{4}^{x}f\left ( t \right )dt$      $\displaystyle 0\leq x\leq 6$
$\displaystyle =3+\int_{4}^{x}\left ( -3 \right )dt=3-3\left ( x-4 \right )=15-3x$
$\displaystyle g\left ( x \right )=0$
$\Rightarrow 15-3x=0$
$\Rightarrow x=5$
$g(x)$ becomes zero at $x=5$
$g(x)$ will be negative in $(5,7)$

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Mathematics

# Value of x at which g (x) becomes zero, is

5

##### SOLUTION
g(x) start decreasing from x=3
$\displaystyle g\left ( 4 \right )=\int_{0}^{4}f\left ( t \right )dt=\int_{0}^{3}f\left ( t \right )dt+\int_{3}^{4}f\left ( t \right )dt$
$\displaystyle =\frac{9}{2}+\int_{3}^{4}\left ( -3t+9 \right )dt=\frac{9}{2}+\left ( 9t+\frac{3t^{2}}{2} \right )^{4}_{3}$
$\displaystyle \frac{9}{2}+\left ( 36-24-27+\frac{27}{2} \right )=3$ Now $\displaystyle g\left ( x \right )=\int_{0}^{x}f\left ( t \right )dt$
$\displaystyle =\int_{0}^{4}f\left ( t \right )dt+\int_{4}^{x}f\left ( t \right )dt$      $\displaystyle 0\leq x\leq 6$
$\displaystyle =3+\int_{4}^{x}\left ( -3 \right )dt=3-3\left ( x-4 \right )=15-3x$
$\displaystyle g\left ( x \right )=0$
$\Rightarrow 15-3x=0$
$\Rightarrow x=5$ which lies in  $[0,6]$

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Mathematics

# Maximum value of g(x) in x $\displaystyle \epsilon$ [0,7] is.

9/2

##### SOLUTION
$\displaystyle g\left ( x \right )=\int_{0}^{x}f\left ( t \right )dt$
$g'(x)=f(x)$ From the graph it is clear that
$\displaystyle f\left ( x \right )> 0$ in $\displaystyle x\epsilon \left [ 0,3 \right ]$ and $\displaystyle f\left ( x \right )< 0$ in $\displaystyle x\epsilon \left [ 3,7 \right ]$
$g(x)$ is increasing in $[0,3]$ and $g(x)$ is decreasing in $[3,7]$
maximum value of $g(x)$ occurs at $x=3$
$\displaystyle g\left ( 3 \right )=\int_{0}^{3}f\left ( t \right )dt$
$\displaystyle =\int_{0}^{1}1\cdot dt+\int_{1}^{2}\left ( 2t-1 \right )dt+\int_{2}^{3}\left ( 3t+9 \right )dt$
$\displaystyle =1+\left ( t^{2}-t \right )^{2}_{1}+\left ( 9t-3\frac{t^{2}}{2} \right )^{3}_{2}$
$=\displaystyle 1+\left ( 4-2-0 \right )+\left ( 27-\frac{27}{2}-18+6 \right )=\frac{9}{2}$

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Questions 203525
Subjects 9
Chapters 126
Enrolled Students 111

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