Find the area bounded by the curve y=x^2+3, the x axis and the ordinates x=0, x=5
To find the area bounded by the curve y=x^2+3, the x-axis, and the ordinates x=0 and x=5, we need to evaluate the definite integral of the curve over the given interval.
The curve y = x^2 + 3 intersects the x-axis when y = 0. Solving the equation x^2 + 3 = 0, we find that there are no real solutions since the minimum value of y = x^2 + 3 is 3, which is greater than 0.
Therefore, the area bounded by the curve, the x-axis, and the ordinates x=0 and x=5 is the area under the curve from x=0 to x=5.
We can calculate this by evaluating the definite integral:
Area = ∫[0, 5] (x^2 + 3) dx
Taking the integral, we get:
Area = [1/3*x^3 + 3x] evaluated from x=0 to x=5
Area = (1/3*5^3 + 3*5) - (1/3*0^3 + 3*0)
Area = (125/3 + 15) - (0/3 + 0)
Area = 170/3
Therefore, the area bounded by the curve y=x^2+3, the x-axis, and the ordinates x=0 and x=5 is 170/3 square units.
To find the area bounded by the curve y=x^2+3, the x-axis, and the ordinates x=0 and x=5, we need to integrate the curve equation within the given limits.
First, let's find the x-coordinate of the points of intersection between the curve and the x-axis.
Setting y=0 in the equation y=x^2+3:
0=x^2+3
x^2=-3
This equation has no real solutions, which means the curve does not intersect the x-axis within the given limits.
Therefore, the area bounded by the curve y=x^2+3, the x-axis, and the ordinates x=0 and x=5 is zero.