Question

I need help .-. Calc

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Answer 1

@AZ

Answer 2

find the eq. of two lines on left of y-axis and right of y-axis.also find the eq. of circle with center (5,3) and radius=3
take f(x)=y
break interval from -2 to 0,0 to 2 and 2 to 5
integrate inthese intervals ,then add you get integration -2 to 5
integration from -6 to 5=integration from -6 to -2 plus integration from -2 to 5
from this you can find your reqd.result for (a)

Answer 3

\[\int\limits_{-6}^{5}f(x)dx=\int\limits_{-6 }^{-2}f(x)dx+\int\limits_{-2}^{-1}f(x)dx+\int\limits_{-1}^{0}(-f(x))dx\]\[+\int\limits_{0}^{0.5}[-f(x)]dx+\int\limits_{0.5}^{2}f(x)dx+\int\limits_{2}^{5}f(x) dx\]

Answer 4

\[\int\limits_{-2}^{-1 }f(x)dx=\frac{ base \times height }{ 2 }=\frac{ 1\times1 }{ 2 }=\frac{ 1 }{ 2 }\]

Answer 5

\[\int\limits_{0}^{0.5}[-f(x) ]dx=\frac{ base \times height }{ 2 }=\frac{ 0.5 \times 1 }{ 2 }=\frac{ 1 }{ 4 }\]

Answer 6

\[\int\limits_{0.5}^{2}f(x) dx=\frac{ base \times height }{ 2 }=\frac{ 1.5 \times 3 }{ 2 }=\frac{ 9 }{ 4 }\]

Answer 7

\[\int\limits_{2}^{5}f(x) dx=3^2-\frac{ \pi \times 3^2}{ 4 }=9-\frac{ 9 \pi }{ 4 }\]
area under the circular arc= area of square-area of quarter circle (r=3)

Answer 8

i have solved without integration.
If you want i can solve with integration also.
Then you have to find the eq. of two lines and circle then evaluate one by one.

Answer 9

now you substitute for all and find the reqd. result.

Answer 10

for b
\[\int\limits_{3}^{5}(2f'(x)+4)dx=2\int\limits_{3}^{5}f'(x) dx+4\int\limits_{3}^{5}dx=2f(x)[3 \rightarrow 5]+4(x)[3 \rightarrow 5]\]
eq. of circle is \[(x-5)^2+(y-3)^2=3^2\]
\[y-3=\sqrt{9-(x-5)^2}\]
when x=3
\[y-3=\sqrt{9-4}\]
\[y=3+\sqrt{5}\]
when x=5
\[y-3=\sqrt{9-0},y=3+3=6\]
so f(x) from 3 to 5 is \[6-(3+\sqrt{5})=3-\sqrt{5}\]
assume y= f(x)

Answer 11

\[\int\limits_{3}^{5}(2 f'(x)+4)dx=2(3-\sqrt{5})+4(5-3)=6-2\sqrt{5}+8=14-2\sqrt{5}\]

Answer 12

I don't get it .-.

Answer 13

let y=f(x) be the curve.
Then
\[\int\limits_{a}^{b}ydx=\int\limits_{a}^{b}f(x) dx\]
It gives the area between ABDC.
it is the area under the curve and x-axis between x=a to x=b

Answer 14

if the curve remains above x-axis then
reqd. area\[=\int\limits_{a}^{b }ydx\]
if it is below x-axis then area\[=\int\limits_{a}^{b}(-y)dx=\int\limits_{a}^{b}(-f(x))dx\]