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Definite Integration question

2022 · Shift 2 · Q21

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JEE Advanced
Mathematics
Definite Integration
2022 · Shift 2 · Q21

Definite Integration question

2022 · Shift 2 · Q21

JEE AdvancedMathematicsDefinite IntegrationNumerical+3 / −1

The greatest integer less than or equal to

\int_{1}^{2} \log _{2}\left(x^{3}+1\right) d x+\int_{1}^{\log _{2} 9}\left(2^{x}-1\right)^{\frac{1}{3}} d x

is

\underline{\hspace{2cm}}

.

Numerical answer

View written solutionFree

Correct answer: 5

Let $I_1=\int_{1}^{2} \log_{2}(x^3+1)\,dx,$ $I_2=\int_{1}^{\log_2 9} (2^x-1)^{1/3}\,dx.$ We need $\left\lfloor I_1+I_2\right\rfloor.$
Rewrite the logarithm in inverse form: $\log_2(x^3+1)=y \iff x=(2^y-1)^{1/3}.$ So the two integrals are of the form $\int f(x)\,dx + \int f^{-1}(x)\,dx.$

Here, $f(x)=\log_2(x^3+1), \qquad f^{-1}(x)=(2^x-1)^{1/3}.$

Use the standard identity for a function and its inverse: If $f(a)=c$ and $f(b)=d$ , then $\int_a^b f(x)\,dx + \int_c^d f^{-1}(y)\,dy = bd-ac.$

For our function: $f(1)=\log_2(2)=1,$ $f(2)=\log_2(9).$ Thus, $I_1+I_2 = 2\log_2 9 - 1.$

Simplify: $\log_2 9=\log_2(3^2)=2\log_2 3,$ so $I_1+I_2=2\log_2 9 -1 = 4\log_2 3 -1.$

Now, $\log_2 3 \approx 1.58496,$ so $I_1+I_2 \approx 4(1.58496)-1 = 6.33984-1 = 5.33984.$

Therefore, $\left\lfloor I_1+I_2\right\rfloor = \lfloor 5.33984\rfloor = 5.$

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