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find the area under the curve (x)=4-(1/4)x^2 over the interval [0,3] with X*k as the right endpoint of each subinterval find the area under the curve (x)=4-(1/4)x^2 over the interval [0,3] with X*k as the right endpoint of each subinterval @Mathematics
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I'm assuming you mean f(x)=4-(1/4)x^2. You'll need to integrate this from x=0 to x=3. \[\int_0^3 4-(1/4)x^2 dx = [4x - (1/12)x^3]_0^3 = (4(3)-(1/12)3^3) - (4(0) - (1/12)0^3) = 12 - 9/4 = 39/4.
i thought something with limits was supposed to happen?
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