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Definite Integration question

2023 · 30 Jan · Shift 1 · Q26
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  5. /2023 · 30 Jan · Shift 1 · Q26

Definite Integration question

2023 · 30 Jan · Shift 1 · Q26

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1
If [t] denotes the greatest integer ≤t\le \mathrm{t}≤t, then the value of 3(e−1)e∫12x2e[x]+[x3]dx{{3(e - 1)} \over e}\int\limits_1^2 {{x^2}{e^{[x] + [{x^3}]}}dx}e3(e−1)​1∫2​x2e[x]+[x3]dx is :
  1. A
    e8−e\mathrm{e^8-e}e8−e
  2. B
    e7−1\mathrm{e^7-1}e7−1
  3. C
    e9−e\mathrm{e^9-e}e9−e
  4. D
    e8−1\mathrm{e^8-1}e8−1
View written solutionFree

Correct answer: A

  1. We need to evaluate
3(e−1)e∫12x2e[x]+[x3] dx.\frac{3(e-1)}{e}\int_1^2 x^2 e^{[x]+[x^3]}\,dx.e3(e−1)​∫12​x2e[x]+[x3]dx.

Here [t][t][t] denotes the greatest integer function.

  1. On the interval [1,2)[1,2)[1,2), we have [x]=1.[x]=1.[x]=1. Also, as xxx goes from 111 to 222, x3x^3x3 goes from 111 to 888. The value of [x3][x^3][x3] changes when x3x^3x3 crosses an integer.

So we split the interval using

for n=1,2,…,7n=1,2,\dots,7n=1,2,…,7.

Thus on [n3,n+13)[\sqrt[3]{n},\sqrt[3]{n+1})[3n​,3n+1​), [x]=1,[x3]=n,[x]=1,\qquad [x^3]=n,[x]=1,[x3]=n, so e[x]+[x3]=e1+n.e^{[x]+[x^3]}=e^{1+n}.e[x]+[x3]=e1+n.

  1. Therefore,
∫12x2e[x]+[x3]dx=∑n=17en+1∫n3n+13x2 dx.\int_1^2 x^2 e^{[x]+[x^3]}dx =\sum_{n=1}^7 e^{n+1}\int_{\sqrt[3]{n}}^{\sqrt[3]{n+1}} x^2\,dx.∫12​x2e[x]+[x3]dx=n=1∑7​en+1∫3n​3n+1​​x2dx.

Now,

∫x2dx=x33.\int x^2 dx = \frac{x^3}{3}.∫x2dx=3x3​.

Hence,

∫n3n+13x2dx=13((n+1)−n)=13.\int_{\sqrt[3]{n}}^{\sqrt[3]{n+1}} x^2 dx =\frac{1}{3}\big((n+1)-n\big)=\frac{1}{3}.∫3n​3n+1​​x2dx=31​((n+1)−n)=31​.

So,

∫12x2e[x]+[x3]dx=13∑n=17en+1=13(e2+e3+⋯+e8).\int_1^2 x^2 e^{[x]+[x^3]}dx =\frac{1}{3}\sum_{n=1}^7 e^{n+1} =\frac{1}{3}(e^2+e^3+\cdots+e^8).∫12​x2e[x]+[x3]dx=31​n=1∑7​en+1=31​(e2+e3+⋯+e8).
  1. This is a geometric series:
e2+e3+⋯+e8=e2(1+e+e2+⋯+e6)=e2⋅e7−1e−1.e^2+e^3+\cdots+e^8 = e^2\left(1+e+e^2+\cdots+e^6\right) = e^2\cdot \frac{e^7-1}{e-1}.e2+e3+⋯+e8=e2(1+e+e2+⋯+e6)=e2⋅e−1e7−1​.

Thus,

∫12x2e[x]+[x3]dx=13⋅e2(e7−1)e−1.\int_1^2 x^2 e^{[x]+[x^3]}dx =\frac{1}{3}\cdot \frac{e^2(e^7-1)}{e-1}.∫12​x2e[x]+[x3]dx=31​⋅e−1e2(e7−1)​.
  1. Multiply by the given factor:
3(e−1)e∫12x2e[x]+[x3]dx=3(e−1)e⋅13⋅e2(e7−1)e−1.\frac{3(e-1)}{e}\int_1^2 x^2 e^{[x]+[x^3]}dx =\frac{3(e-1)}{e}\cdot \frac{1}{3}\cdot \frac{e^2(e^7-1)}{e-1}.e3(e−1)​∫12​x2e[x]+[x3]dx=e3(e−1)​⋅31​⋅e−1e2(e7−1)​.

Canceling,

=e(e7−1)=e8−e.= e(e^7-1)=e^8-e.=e(e7−1)=e8−e.
  1. Hence the correct option is A  :  e8−e.\boxed{A\;:\; e^8-e}.A:e8−e​.
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