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∫[x 2 ]dx limit is from 0 to 2.

∫[x2]dx
limit is from 0 to 2.

Grade:12

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
6 years ago
Ans:
Hello Student,
Please find answer to your question below

I = \int_{0}^{2}[x^2]dx
I = \int_{0}^{1}[x^2]dx+\int_{1}^{\sqrt{2}}[x^2]dx+\int_{\sqrt{2}}^{\sqrt{3}}[x^2]dx+\int_{\sqrt{3}}^{2}[x^2]dx
I = \int_{0}^{1}(0)dx+\int_{1}^{\sqrt{2}}(1)dx+\int_{\sqrt{2}}^{\sqrt{3}}[x^2]dx+\int_{\sqrt{3}}^{2}[x^2]dx
I = (\sqrt{2}-1)+\int_{\sqrt{2}}^{\sqrt{3}}[x^2]dx+\int_{\sqrt{3}}^{2}[x^2]dx
I = (\sqrt{2}-1)+\int_{\sqrt{2}}^{\sqrt{3}}(2)dx+\int_{\sqrt{3}}^{2}[x^2]dx
I = (\sqrt{2}-1)+2(\sqrt{3}-\sqrt{2})+\int_{\sqrt{3}}^{2}(3)dx
I = (\sqrt{2}-1)+2(\sqrt{3}-\sqrt{2})+3(2-\sqrt{3})
I = 5-\sqrt{2}-\sqrt{3}

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