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Joint Entrance Examination
Mathematics
Definite Integration
Question 191
Question 191 of 246
Medium
MCQ
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 :
A
\[-\]1
B
\[{{ - 5} \over 4}\]
C
\[{{\sqrt {17} - 13} \over 8}\]
D
\[{{\sqrt {17} - 16} \over 8}\]
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Official solution
∫
0
1
[
−
8
x
2
+
6
x
−
1
]
d
x
\int_{0}^{1}\left[-8 x^{2}+6 x-1\right] d x
∫
0
1
[
−
8
x
2
+
6
x
−
1
]
d
x
=
∫
0
1
4
(
−
1
)
d
x
+
∫
1
4
3
4
0
d
x
+
∫
1
2
3
4
−
1
d
x
+
∫
3
4
8
−
2
d
x
+
∫
3
+
17
8
1
−
3
d
x
=\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
=
∫
0
4
1
(
−
1
)
d
x
+
∫
4
1
4
3
0
d
x
+
∫
2
1
4
3
−
1
d
x
+
∫
4
3
8
−
2
d
x
+
∫
8
3
+
17
1
−
3
d
x
=
−
1
4
−
1
4
−
2
(
3
+
17
8
−
3
4
)
−
3
(
1
−
3
+
17
8
)
=-\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)
=
−
4
1
−
4
1
−
2
(
8
3
+
17
−
4
3
)
−
3
(
1
−
8
3
+
17
)
=
17
−
13
8
=\frac{\sqrt{17}-13}{8}
=
8
17
−
13
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