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Q15E

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Essential Calculus: Early Transcendentals
Found in: Page 289
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

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Short Answer

Evaluate the integral.

\(\int_{\rm{0}}^{\rm{1}} {\left( {{{\rm{x}}^{{\rm{10}}}}{\rm{ + 1}}{{\rm{0}}^{\rm{x}}}} \right)} {\rm{dx}}\).

The solution of given integral\(\int_0^1 {\left( {{x^{10}} + {{10}^x}} \right)} dx\) is \(\frac{1}{{11}} + \frac{9}{{\ln 10}}\).

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Power rule of integration

Power rule of integration allows us to find the indefinite and definite integrals of a variety of functions like polynomials, functions involving roots and even rational functions.

\(\int_a^b {{x^n}} dx = \left. {\frac{{{x^{n + 1}}}}{{n + 1}}} \right|_a^b\) and \(\int_b^c {{a^x}} dx = \left. {\frac{{{a^x}}}{{\ln a}}} \right|_b^c\)

Step 2: Applying power rule of integration

Using evaluation theorem write given integral as-

\(\begin{aligned}{c}\int_0^1 {{x^{10}}} + {10^x}dx &= \left( {\frac{{{x^{10 + 1}}}}{{10 + 1}} + \frac{{{{10}^x}}}{{\ln 10}}} \right)_0^1\\ &= \left( {\frac{{{x^{11}}}}{{11}} + \frac{{{{10}^x}}}{{\ln 10}}} \right)_0^1\\ &= \left( {\frac{{{1^{11}}}}{{11}} + \frac{{{{10}^1}}}{{\ln 10}}} \right) - \left( {\frac{{{0^{11}}}}{{11}} + \frac{{{{10}^0}}}{{\ln 10}}} \right)\\ &= \left( {\frac{1}{{11}} + \frac{{10}}{{\ln 10}}} \right) - \left( {0 + \frac{1}{{\ln 10}}} \right)\\ &= \frac{1}{{11}} + \frac{9}{{\ln 10}}\end{aligned}\)

Therefore, the solution of given integral\(\int_0^1 {\left( {{x^{10}} + {{10}^x}} \right)} dx\) is \(\frac{1}{{11}} + \frac{9}{{\ln 10}}\).

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