If ( I_ m =∫_ 1^ e ( x)^ m dx ), where ( m ∈ N ), then I_ 10 +10 I_9 is equal to -

Correct answer: C. ( e ) ( aligned & I_ 10 = _1^ e 1 ·( x)^ 10 dx=[( x)^ 10 x ]_1^ e & -∫_1^ e 10( x)^9 · 1 x · xdx=e-0-10 ∫_1^ e ( x)^9 dx & =e-10 I _ 9 +10 I _9=e aligned )

BITSAT 2010

Exam: BITSAT
Year: 2010
Subject: Mathematics
Chapter: Area Under Curves

BITSAT 2010 · Mathematics · Area Under Curves

If ( I_ m =∫_ 1^ e ( x)^ m dx ), where ( m ∈ N ), then I_ 10 +10...

If (

\mathrm{I}_{\mathrm{m}}=\int_

1^{

\mathrm{e}

}(\ln

\mathrm{x})^{\mathrm{m}} \mathrm{dx}

), where (

\mathrm{m} \in \mathrm{N}

), then

I_{10}+10 I_9

is equal to -

Options

A(\mathrme^{10})
B( $\frac{\mathrme^{10}}{10}$ )
C( $\mathrm{e}$ )
D $\mathrm{e}-1)$

Correct answer

C. ( $\mathrm{e}$ )

Step-by-step solution

(

\begin{aligned} & \mathrm{I}_{10}=\iint_1^{\mathrm{e}}1 \cdot(\ln \mathrm{x})^{10} \mathrm{dx}=\left[(\ln \mathrm{x})^{10} \mathrm{x}\rright]_1^{\mathrm{e}}\\ & -\int_1^{\mathrm{e}}10(\ln \mathrm{x})^9 \cdot \frac{1}{\mathrm{x}} \cdot \mathrm{xdx}=\mathrm{e}-0-10 \int_1^{\mathrm{e}}(\ln \mathrm{x})^9 \mathrm{dx}\\ & =\mathrm{e}-10 \mathrm{I}}_{9}+10 \mmathrm{I}_9=\mathrm{e} \end{aligned}

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