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
❖ᴏɴʟʏ ꜰᴏʀ ᴍᴏᴅᴇʀᴀᴛᴏʀꜱ, ʙʀᴀɪɴʟʏ ꜱᴛᴀʀꜱ ᴀɴᴅ ᴏᴛʜᴇʀ ʙᴇꜱᴛ ᴜꜱᴇʀꜱ​​​​​​​​​

Answer 1

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 trapezoidal approximation found with 3 sub intervals is 52.

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

We,know that The formula for the area of a trapezoid is

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

Where h = delta (x)

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

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

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

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

$\Rightarrow 32 + 2k = 52$

$\Rightarrow 2k = 52 - 32$

$\Rightarrow 2k = 20$

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

$\Rightarrow k =10$

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

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