Question

Which of the following functions has the highest area under the curve between \(x=0\) and \(x=10\)?

• A. \(y=x+8\)
• B. \(y=3x\)
• C. \(y=2x+1\)
• D. \(y=x+2\)

Hint:

Compute the definite integral (or the trapezoid area) of each line from x=0 to x=10 and compare the four values.

Answer 1

The Correct Option is B

Step 1: Understanding the Question.
We must find, among the four given straight lines, which one has the largest area under it between \(x=0\) and \(x=10\).

Step 2: Key Formula or Approach.
The area under a curve \(y=f(x)\) between \(x=a\) and \(x=b\) is the definite integral \(\int_a^b f(x)\,dx\). We will integrate each option from \(x=0\) to \(x=10\).

Step 3: Detailed Explanation.
Option (A): \(y = x+8\).
\[ \int_0^{10} (x+8)\,dx = \left[\frac{x^2}{2}+8x\right]_0^{10} = \left(\frac{100}{2}+80\right) - 0 = 50+80 = 130 \]
Option (B): \(y=3x\).
\[ \int_0^{10} 3x\,dx = \left[\frac{3x^2}{2}\right]_0^{10} = \frac{3(100)}{2} = 150 \]
Option (C): \(y=2x+1\).
\[ \int_0^{10}(2x+1)\,dx = \left[x^2+x\right]_0^{10} = 100+10 = 110 \]
Option (D): \(y=x+2\).
\[ \int_0^{10}(x+2)\,dx = \left[\frac{x^2}{2}+2x\right]_0^{10} = 50+20 = 70 \]
Comparing all four areas, 130, 150, 110, and 70, the largest is 150, from option (B).

Step 4: Final Answer.
Option (B), \(y=3x\), gives the highest area under the curve, 150 square units, between \(x=0\) and \(x=10\). \[ \boxed{y=3x} \]