JEE MAIN - Mathematics (2022 - 28th June Morning Shift - No. 5)
Let [t] denote the greatest integer less than or equal to t. Then, the value of the integral $$\int\limits_0^1 {[ - 8{x^2} + 6x - 1]dx} $$ is equal to :
$$-$$1
$${{ - 5} \over 4}$$
$${{\sqrt {17} - 13} \over 8}$$
$${{\sqrt {17} - 16} \over 8}$$
Explanation
$\int_{0}^{1}\left[-8 x^{2}+6 x-1\right] d x$
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$=\int_{0}^{\frac{1}{4}}(-1) d x+\int_{\frac{1}{4}}^{\frac{3}{4}} 0 d x+\int_{\frac{1}{2}}^{\frac{3}{4}}-1 d x+\int_{\frac{3}{4}}^{8}-2 d x+\int_{\frac{3+\sqrt{17}}{8}}^{1}-3 d x$
$=-\frac{1}{4}-\frac{1}{4}-2\left(\frac{3+\sqrt{17}}{8}-\frac{3}{4}\right)-3\left(1-\frac{3+\sqrt{17}}{8}\right)$
$=\frac{\sqrt{17}-13}{8}$
$=\int_{0}^{\frac{1}{4}}(-1) d x+\int_{\frac{1}{4}}^{\frac{3}{4}} 0 d x+\int_{\frac{1}{2}}^{\frac{3}{4}}-1 d x+\int_{\frac{3}{4}}^{8}-2 d x+\int_{\frac{3+\sqrt{17}}{8}}^{1}-3 d x$
$=-\frac{1}{4}-\frac{1}{4}-2\left(\frac{3+\sqrt{17}}{8}-\frac{3}{4}\right)-3\left(1-\frac{3+\sqrt{17}}{8}\right)$
$=\frac{\sqrt{17}-13}{8}$
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