Area under a line

Calculus Level 1

What is the area between y=3x+3 and the x-axis from x-values 5 to 10?

255/2 130 255/4 678/3

3 solutions

Brock Brown
Sep 28, 2015

Here's the geometric solution:

A calculus solution could be $\int_5^{10} 3x + 3\ dx = [\frac{3}{2}(10)^2+3(10)]-[\frac{3}{2}(5)^2+3(5)] = \boxed{\frac{255}{2}}$ .

CHALLENGE MASTER NOTE: How about from y = 5 to y = 10? How would you set up the integral?

Pi Han Goh - 5 years, 8 months ago

Just rephrase the original function in terms of x instead of terms of y. Then evaluate it like you would any other integral.

$y = 3x + 3 \implies$

$x = \frac{y}{3} - 1 \implies$

$\int_{5}^{10} \frac{y}{3} - 1 \ dy = (\frac{10^2}{6}-10) - (\frac{5^2}{6}-5) = \frac{75}{6} - 5 = \boxed{7.5}$

Brock Brown - 5 years, 8 months ago

Protip: Write the indefinite integral first, so you don't have to recall what it is when you do the diferrence.

$\int_5^{10} \frac y3 - 1 \, dy = \left[ \frac{y^2}6 - y\right]_5^{10} = \ldots$

Pi Han Goh - 5 years, 8 months ago

Aaaaa Bbbbb
Sep 28, 2015

It Is very to realize that the x-axis and two lines: x=5 and y=10 make a trapezoid that have area is: $A = \frac{(18+33) \times 5}{2}=\boxed{\frac{255}{2}}$

Arjen Vreugdenhil
Oct 11, 2015

Because the graph is a straight line, we can find the area under the graph simply by considering the midpoint, $x = 15/2$ : $A = 5 \times \left(3\cdot \frac{15}{2} + 3\right) = 5 \times \frac{51}{2} = \boxed{\frac{255}{2}}.$

Area under a line

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