I do not know how to work this problem:
Find the area of the region between the graph of y=x^3 and the x-axis from x=1 to x=4.
I am also having trouble with: Write
-1-i in exponential form.
Any help would be greatly appreciated.
The heght of the curve is x^3 each incremental area then is height*width, x^3 dx
INT x^3 dx from x=1 to 4
I am still lost
To find the area of the region between the graph of y = x^3 and the x-axis from x = 1 to x = 4, you can use the definite integral. The area can be calculated by integrating the absolute value of the function y = x^3 over the given range of x values.
Here are the steps to solve this problem:
1. Set up the integral: ∫[1, 4] |x^3| dx
This means you integrate the absolute value of x^3 over the interval from x = 1 to x = 4.
2. Split the integral into two parts: ∫[1, 4] x^3 dx + ∫[1, 4] -x^3 dx
Since x^3 is positive over the given range, it remains as it is in the first part. In the second part, since the function is below the x-axis, you change the sign to negative.
3. Evaluate each integral separately:
∫[1, 4] x^3 dx = [(1/4) x^4] [1, 4] = (1/4) * (4^4) - (1/4) * (1^4) = 63/4
∫[1, 4] -x^3 dx = -∫[1, 4] x^3 dx = -63/4
4. Add the results from the two parts: (63/4) + (-63/4) = 0
Since the two integrals cancel each other out due to their opposite signs, the resulting area is zero.
Now let's move on to the second problem:
To write -1 - i in exponential form (a + bi), you can use Euler's formula, which states that e^(iθ) = cos(θ) + i sin(θ), where θ is the angle in radians.
1. Rewrite -1 - i as a complex number in the form a + bi:
-1 - i = -1 + (-1)i
2. Convert it to polar form using the fact that r = sqrt(a^2 + b^2) and θ = arctan(b/a):
r = sqrt((-1)^2 + (-1)^2) = sqrt(2)
θ = arctan((-1)/(-1)) = arctan(1) = π/4 (45 degrees or π/4 radians)
3. Write it in exponential form:
-1 - i = sqrt(2) * e^(iπ/4)
This is the exponential form of -1 - i. Notice that the length of the vector (sqrt(2)) is represented by the magnitude r, and the angle π/4 is represented by the argument θ.
I hope this helps! Let me know if you have any further questions.