KCET · Maths · Definite Integration
If \([x]\) is the greatest integer function not greater than \(x\), then \(\int_0^8[x] d x\) is equal to
- A 28
- B 30
- C 29
- D 20
Answer & Solution
Correct Answer
(A) 28
Step-by-step Solution
Detailed explanation
Let \(I=\int_0^8[x] d x\)
\[
\begin{aligned}
& I=\int_0^1[x] d x+\int_1^2[x] d x+\int_2^3[x] d x+\int_3^4[x] d x+\int_4^5[x] d x \\
& +\int_5^6[x] d x+\int_6^7[x] d x+\int_7^8[x] d x \\
& =\int_0^1 0 d x+\int_1^2 1 d x+\int_2^3 2 d x+\int_3^4 3 d x+\int_4^5 4 d x \\
& +\int_5^6 5 d x+\int_6^7 6 d x+\int_7^8 7 d x \\
& =0+1[x]_1^2+2[x]_2^3+3[x]_3^4+4[x]_4^5 \\
& +5[x]_5^6+6[x]_6^7+7[x]_7^8 \\
& =0+1(2-1)+2(3-2)+3(4-3) \\
& +4(5-4)+5(6-5)+6(7-6)+7(8-7) \\
& =1+2+3+4+5+6+7 \\
& =\frac{7 \cdot(7+1)}{2}=\frac{7 \cdot 8}{2}=7 \cdot 4=28 \text {. } \\
&
\end{aligned}
\]
\[
\begin{aligned}
& I=\int_0^1[x] d x+\int_1^2[x] d x+\int_2^3[x] d x+\int_3^4[x] d x+\int_4^5[x] d x \\
& +\int_5^6[x] d x+\int_6^7[x] d x+\int_7^8[x] d x \\
& =\int_0^1 0 d x+\int_1^2 1 d x+\int_2^3 2 d x+\int_3^4 3 d x+\int_4^5 4 d x \\
& +\int_5^6 5 d x+\int_6^7 6 d x+\int_7^8 7 d x \\
& =0+1[x]_1^2+2[x]_2^3+3[x]_3^4+4[x]_4^5 \\
& +5[x]_5^6+6[x]_6^7+7[x]_7^8 \\
& =0+1(2-1)+2(3-2)+3(4-3) \\
& +4(5-4)+5(6-5)+6(7-6)+7(8-7) \\
& =1+2+3+4+5+6+7 \\
& =\frac{7 \cdot(7+1)}{2}=\frac{7 \cdot 8}{2}=7 \cdot 4=28 \text {. } \\
&
\end{aligned}
\]
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