Question

The function f is continuous on the closed interval [0,6] and has values given in the table above. The trapezoidal approximation found with 3 subintervals is 52. What is the value of k?

❖ᴏɴʟʏ ᴘʀᴏᴘᴇʀ ꜱᴏʟᴠᴇᴅ ᴀɴꜱᴡᴇʀ ᴡɪᴛʜ ɢᴏᴏᴅ ᴇxᴘʟᴀɴᴀɪᴏɴ ɴᴇᴇᴅᴇᴅ
❖ ɴᴏ ꜱᴘᴀᴍᴍɪɴɢ
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Answer 1

Answer:

The function f is continuous on the closed interval [0,6] and has values that are given in the table below.

x |0|2|4|6

f(x)|4|K|8|12

The trapezoidal approximation for(the integral):

6

S f(x) dx

1

found with 3 subintervals of equal length is 52. What is the value of K?

How do you do this with the formula

b-a/(2n) * f(x0) + 2(fx1) + f(xn)?

The formula for the area of a trapezoid is 1/2(b_1 + b_2)h where h = delta (x)

Therefore, A = 1/2(4+k)2 + 1/2(k+8)2 + 1/2(8+12)2 = 54

Solving that equation yields k = 10.

Answer 2

Answer:

The value of k =10

Step-by-step explanation:

$\green{\ggg}\red{\underline{Question:-}}$

The function f is continuous on the closed interval [0,6] and has values given in the table above. The trapezoidal approximation found with 3 sub intervals is 52. What is the value of k?

$\green{\ggg} \red{\underline{Solution:-}}$

They given,

The function f is continuous on the closed interval [0,6]

The trapezoidal approximation found with 3 sub intervals is 52.

We know,the formula of trapezoidal

$:\longrightarrow \boxed{\frac{1}{2} (b_1 +b_2)h}$

$\bold{Therefore,A=} \frac{1}{2}(4+k)2+\frac{1}{2}(k+8)2+\frac{1}{2} (8+12)2 = 52$

$\Rightarrow \frac{1}{2}(4+k)2+\frac{1}{2}(k+8)2+\frac{1}{2} (8+12)2 = 52$

$\Rightarrow( 4 + k)+(k+8)+(8+12)=52$

$\Rightarrow 12+2k + 20 = 52$

$\Rightarrow 32 + 2k = 52$

$\Rightarrow 2k = 52 - 32$

$\Rightarrow 2k = 20$

$\Rightarrow k =\frac{20}{2}$

$\Rightarrow k =10$

$\pink{:\longrightarrow}\boxed{\red{k=10}}$

The equation yields k = 10 and the value of k =10