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 = β«[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.
Find the area bounded by the curve y=x^2+3, the x axis and the ordinates x=0, x=5
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