Integral Calculus> ∫[x 2 ]dx limit is from 0 to 2....

∫[x²]dx
limit is from 0 to 2.

Lovey , 11 Years ago

Grade 12

1 Answers

Jitender Singh

Last Activity: 11 Years ago

Ans:
Hello Student,
Please find answer to your question below

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