11th Standard Syllabus & Materials

11th Standard

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11th Standard English Medium Maths Subject Integral Calculus Book Back 2 Mark Questions with Solution Part - I

Published on: 07/06/2021

QB365 provides detailed and simple solution for every book back questions in class 11 Maths subject.It will helps to get more idea about question pattern in every book back questions with solution.

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Std: 11th Standard - Maths

chapter: - Book back Questions

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Questions:

Generic 2

10 x 2 = 20

Integrate the following with respect to x : \(\sqrt [3]{x^4}\)

Integrate the following with respect to x : \({1\over x^7}\)

Integrate the following with respect to x : \({1\over 1+x^2}\)

Integrate the following with respect to x :\({1\over x^3}\)

Integrate the following with respect to x : \({1\over e^{-x}}\)

Integrate the following with respect to x : \({1\over \sqrt{1-x^2}}\)

Integrate the following with respect to x : \({sin \ x \over cos ^2 \ x}\)

IIntegrate the following with respect to x : \({cot \ x \over sin \ x}\)

Integrate the following with respect to x : \(\sqrt{x}\)

10.

Integrate the following with respect to x : x¹⁰

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Answers Key:

Generic 2

10 x 2 = 20

\( \int \sqrt[3]{x^4} d x =\int\left(x^4\right)^{1 / 3} d x \)
\( =\int x^{\frac{4}{3}} d x \)
\(=\frac{x^{\frac{4}{3}+1}}{\frac{4}{3}+1}+c=\frac{3}{7} x^{\frac{7}{3}}+c \)

\(\int \frac{1}{x^7} d x=\frac{x^{-7+1}}{-7+1}+c=\frac{x^{-6}}{-6}+c=-\frac{1}{6 x^6}+c\)

\(\int {1\over 1+x^2}dx=tan^{-1}x+c\)

\(\int{1\over x^3}dx=\int{x^{-3}dx}={x^{-3+1}\over -3+1}+c=-{1\over 2x^2}+c\)

\(\int{1\over e^{-x}}dx=\int{e^x }dx=e^x+c\)

\(\int{{1\over \sqrt{1-x^2}}dx}=sin ^{-1}x+c\)

\(\int{ {sin \ x \over cos^2 \ x}dx}= \int{{sin \ x\over cos \ x}.{1\over cos \ x} \ dx}= \int {tan \ x sec \ x \ dx \ = sec \ x +c}\)

\(\int \frac{\cot x}{\sin x} d x=\int \operatorname{cosec} x \cot xd x=-\operatorname{cosec} x+c\)

\(\int {\sqrt{x}\ dx }= \int { {x^{1\over 2}} }dx={x^{{1\over 2}+1}\over {1\over2}+1}+c={x^{3\over 2}\over {3\over2}}+c={2\over3}x^{3\over2}+c\)

10.

We know that \(\int { x^n \ dx }={x^{n+1}\over n+1} +c,n\neq -1.\)
Putting n = 10, we get
\(\int { x^{10} \ dx }={x^{10+1}\over 10+1} +c={x^{11}\over 11}+c\)