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If f is continuously differentiable func...

If `f` is continuously differentiable function then `int_(0)^(2.5) [x^2] f'(x) dx` is equal to

A

`6f(6.25)+ underset(i in {1, sqrt2, sqrt3, 2, sqrt5})(sumf (i))`

B

`6f(6.25)- underset(i in {1, sqrt2, sqrt3, 2, sqrt5, sqrt6})(sumf (i))`

C

`6f(2.5)- underset(i in {1, sqrt2, sqrt3, 2, sqrt5, sqrt6})(sumf (i))`

D

`6f(2.5)- underset(i in {1, sqrt2, sqrt3, sqrt5})(sumf (i))`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the integral \( \int_{0}^{2.5} \lfloor x^2 \rfloor f'(x) \, dx \), where \( f \) is a continuously differentiable function, we need to break down the integral based on the behavior of the greatest integer function \( \lfloor x^2 \rfloor \). ### Step 1: Identify intervals for \( \lfloor x^2 \rfloor \) The function \( \lfloor x^2 \rfloor \) takes integer values depending on the range of \( x \): - For \( x \in [0, 1) \), \( \lfloor x^2 \rfloor = 0 \) - For \( x \in [1, \sqrt{2}) \), \( \lfloor x^2 \rfloor = 1 \) - For \( x \in [\sqrt{2}, \sqrt{3}) \), \( \lfloor x^2 \rfloor = 2 \) - For \( x \in [\sqrt{3}, 2) \), \( \lfloor x^2 \rfloor = 3 \) - For \( x \in [2, \sqrt{5}) \), \( \lfloor x^2 \rfloor = 4 \) - For \( x \in [\sqrt{5}, \sqrt{6}) \), \( \lfloor x^2 \rfloor = 5 \) - For \( x \in [\sqrt{6}, 2.5] \), \( \lfloor x^2 \rfloor = 6 \) ### Step 2: Break the integral into parts Thus, we can break the integral into the following segments: \[ \int_{0}^{2.5} \lfloor x^2 \rfloor f'(x) \, dx = \int_{0}^{1} 0 \cdot f'(x) \, dx + \int_{1}^{\sqrt{2}} 1 \cdot f'(x) \, dx + \int_{\sqrt{2}}^{\sqrt{3}} 2 \cdot f'(x) \, dx + \int_{\sqrt{3}}^{2} 3 \cdot f'(x) \, dx + \int_{2}^{\sqrt{5}} 4 \cdot f'(x) \, dx + \int_{\sqrt{5}}^{\sqrt{6}} 5 \cdot f'(x) \, dx + \int_{\sqrt{6}}^{2.5} 6 \cdot f'(x) \, dx \] ### Step 3: Evaluate each integral 1. **First Integral**: \[ \int_{0}^{1} 0 \cdot f'(x) \, dx = 0 \] 2. **Second Integral**: \[ \int_{1}^{\sqrt{2}} f'(x) \, dx = f(\sqrt{2}) - f(1) \] 3. **Third Integral**: \[ \int_{\sqrt{2}}^{\sqrt{3}} 2 f'(x) \, dx = 2(f(\sqrt{3}) - f(\sqrt{2})) \] 4. **Fourth Integral**: \[ \int_{\sqrt{3}}^{2} 3 f'(x) \, dx = 3(f(2) - f(\sqrt{3})) \] 5. **Fifth Integral**: \[ \int_{2}^{\sqrt{5}} 4 f'(x) \, dx = 4(f(\sqrt{5}) - f(2)) \] 6. **Sixth Integral**: \[ \int_{\sqrt{5}}^{\sqrt{6}} 5 f'(x) \, dx = 5(f(\sqrt{6}) - f(\sqrt{5})) \] 7. **Seventh Integral**: \[ \int_{\sqrt{6}}^{2.5} 6 f'(x) \, dx = 6(f(2.5) - f(\sqrt{6})) \] ### Step 4: Combine the results Now, combining all the results, we have: \[ \int_{0}^{2.5} \lfloor x^2 \rfloor f'(x) \, dx = 0 + (f(\sqrt{2}) - f(1)) + 2(f(\sqrt{3}) - f(\sqrt{2})) + 3(f(2) - f(\sqrt{3})) + 4(f(\sqrt{5}) - f(2)) + 5(f(\sqrt{6}) - f(\sqrt{5})) + 6(f(2.5) - f(\sqrt{6})) \] This simplifies to: \[ = 6f(2.5) - f(1) - f(\sqrt{2}) - f(\sqrt{3}) - f(2) - f(\sqrt{5}) - f(\sqrt{6}) \] ### Final Answer Thus, the value of the integral is: \[ \int_{0}^{2.5} \lfloor x^2 \rfloor f'(x) \, dx = 6f(2.5) - (f(1) + f(\sqrt{2}) + f(\sqrt{3}) + f(2) + f(\sqrt{5}) + f(\sqrt{6})) \]
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