find the area between the curve y= x^2+4x and the x - axis from x=-2 to x= 0.
Area under a curve - region bounded by the given function, vertical
lines and the x axis =
a
∫ y (x) dx
b
In this case:
∫ ( x² + 4 x ) dx = x³ / 3 + 4 x² / 2 = x³ / 3 + 2 x² + C
Area =
2
∫ ( x² + 4 x ) dx =
0
0
| x³ / 3 + 2 x² | = F(0) - F(2) =
- 2
0³ / 3 + 2 ∙ 0² - [ ( - 2 )³ / 3 + 2 ∙ ( - 2 )² ] = 0 - ( - 8 / 3 + 2 ∙ 4 ) =
0 - ( - 8 / 3 + 8 ) = 0 - ( - 8 / 3 + 24 / 3 ) = 0 - ( 16 / 3 ) = - 16 / 3
A negative area value means the area is below the x-axis
The area cannot be negative, so the area value is:
A = | - 16 / 3 |
A = 16 / 3
My typo.
Area is not
2
∫ ( x² + 4 x ) dx
0
Area =
0
∫ ( x² + 4 x ) dx =
- 2
Sure! Let's find the area between the curve y = x^2 + 4x and the x-axis from x = -2 to x = 0.
To find the area, we need to integrate the absolute value of the curve equation between the given x-values.
Since the curve is below the x-axis in the given interval, we need to take the absolute value to ensure a positive area.
The integral for the area is:
∫[from -2 to 0] |x^2 + 4x| dx
Let's integrate this and find the area:
∫[from -2 to 0] |x^2 + 4x| dx
= ∫[from -2 to 0] (x^2 + 4x) dx
Oh, wait! Sorry, my calculations are a little fuzzy today. Let me reconsider.
Hmm, instead of doing all that math, let me tell you a joke to lighten the mood! What do you call a fish wearing a crown? King Cod! Now, let's get back to the problem.
To find the area, we need to break it into two parts since the curve intersects the x-axis at x = -2. Let's start with the part between x = -2 and x = 0.
First, we integrate the equation y = x^2 + 4x:
∫[from -2 to 0] (x^2 + 4x) dx
Integrating this will give us the area between the curve and the x-axis in that range. Let's calculate:
∫[from -2 to 0] (x^2 + 4x) dx
= [1/3 * x^3 + 2x^2] ∣[from -2 to 0]
Now, let's plug in the values and evaluate the expression:
[1/3 * 0^3 + 2 * 0^2] - [1/3 * (-2)^3 + 2 * (-2)^2]
= [0 + 0] - [(-8/3) + 8]
= 8 - 8/3
= 24/3 - 8/3
= 16/3
So, the area between the curve y = x^2 + 4x and the x-axis from x = -2 to x = 0 is 16/3 square units.
If you have any more questions or need further assistance, feel free to ask! I'm here to help, and maybe crack a joke or two along the way!
To find the area between the curve y = x^2 + 4x and the x-axis from x = -2 to x = 0, we need to integrate the function over that interval.
Step 1: Determine the integral expression for the given function (y = x^2 + 4x) over the given interval (x = -2 to x = 0).
The given function is y = x^2 + 4x, and the interval is from x = -2 to x = 0. We can write the integral expression as follows:
A = ∫[from -2 to 0] (x^2 + 4x) dx
Step 2: Integrate the function over the given interval.
To integrate the function (x^2 + 4x), we can split it into two separate integrals:
A = ∫[from -2 to 0] x^2 dx + ∫[from -2 to 0] 4x dx
Let's solve each integral separately:
∫ x^2 dx:
Integrate x^2 with respect to x by adding 1 to the exponent and dividing by the new exponent:
∫ x^2 dx = (1/3) x^3
∫ 4x dx:
Integrate 4x with respect to x by adding 1 to the exponent and dividing by the new exponent:
∫ 4x dx = 2x^2
Step 3: Plug in the limits of integration and evaluate the integrals.
To find the area, we need to evaluate each integral at the upper limit (x = 0) and subtract the result from the evaluation at the lower limit (x = -2).
A = [(1/3) (0)^3 - (1/3) (-2)^3] + [2(0)^2 - 2(-2)^2]
A = [(1/3) (0) - (1/3) (-8)] + [0 - 2(4)]
A = [0 + (8/3)] + [0 - 8]
A = (8/3) - 8
A = -16/3
Step 4: Determine the absolute value of the area.
Since area cannot be negative, we need to take the absolute value of the result:
|A| = |-16/3|
|A| = 16/3
Therefore, the area between the curve y = x^2 + 4x and the x-axis from x = -2 to x = 0 is 16/3 square units.
Recall the Fundamental Theorem of Calculus and Riemann sums. You just want
∫[-2,0] (0 - (x^2 + 4x)) dx = 16/3