Question

The graph of f is given in the figure to the right. Let ​A(x)equals=Integral from 0 to x f left parenthesis t right parenthesis font size decreased by 3 dt∫0xf(t) dt and evaluate ​A(4​), ​A(8​), ​A(12​), and ​A(14)

Answer 1

Answer:

$A(4)=-4\pi$

$A(8)=-4\pi +8$

$A(12)=-4\pi +16$

$A(14)=-4\pi +15$

Step-by-step explanation:

we are given

$A(x)=\int\limits^x_0 f{x} \, dx$

Calculation of A(4):

we can plug x=4

$A(4)=\int\limits^4_0 f{x} \, dx$

Since, this curve is below x-axis

so, the value of integral must be negative

and it is quarter of circle

so, we can find area of quarter circle

radius =4

$A(4)=-\frac{1}{4}\times \pi \times (4)^2$

$A(4)=-4\pi$

Calculation of A(8):

we can plug x=8

$A(8)=\int\limits^8_0 f{x} \, dx$

we can break into two parts

$A(8)=\int\limits^4_0 f{x} \, dx+\int\limits^8_4 f{x} \, dx$

now, we can find area and then combine them

$A(8)=-4\pi +\frac{1}{2}\times 4\times 4$

$A(8)=-4\pi +8$

Calculation of A(12):

we can plug x=12

$A(12)=\int\limits^12_0 f{x} \, dx$

we can break into two parts

$A(12)=\int\limits^4_0 f{x} \, dx+\int\limits^8_4 f{x} \, dx+\int\limits^12_8 f{x} \, dx$

now, we can find area and then combine them

$A(12)=-4\pi +\frac{1}{2}\times 8\times 4$

$A(12)=-4\pi +16$

Calculation of A(14):

we can plug x=14

$A(14)=\int\limits^14_0 f{x} \, dx$

we can break into two parts

$A(14)=\int\limits^4_0 f{x} \, dx+\int\limits^8_4 f{x} \, dx+\int\limits^12_8 f{x} \, dx+\int\limits^14_12 f{x} \, dx$

now, we can find area and then combine them

$A(14)=-4\pi +\frac{1}{2}\times 8\times 4-\frac{1}{2}\times 1\times 2$

$A(14)=-4\pi +16-1$

$A(14)=-4\pi +15$