Ex 7.7 class 12 maths ncert solutions | class 12 maths exercise 7.7 | class 12 maths ncert solutions chapter 7 exercise 7.7 | exercise 7.7 class 12 maths ncert solutions | integrals class 12 ncert solutions
Exercise 7.7 Class 12 Maths NCERT Solutions focuses on definite integrals, introducing students to the concept of evaluating integrals within given limits. The problems in Class 12 Maths Exercise 7.7 help strengthen understanding of how integration applies to real numerical intervals. Our Class 12 Maths NCERT Solutions Chapter 7 Exercise 7.7 are designed to explain each question clearly and step-by-step. These Exercise 7.7 Class 12 Maths NCERT Solutions are based on the CBSE syllabus and form a key part of the Integrals Class 12 NCERT Solutions, which are crucial for board and entrance exam preparation.
exercise 7.7 class 12 maths ncert solutions || class 12 maths ncert solutions chapter 7 exercise 7.7 || integrals class 12 ncert solutions || class 12 maths exercise 7.7 || ex 7.7 class 12 maths ncert solutions
Exercise 7.7
1. Integrate: \(\sqrt{4-x^{2}}\)
Answer
\(\int \sqrt{4-x^{2}} d x=\int \sqrt{(2)^{2}-(x)^{2}} d x\)
We know that,
\(\Rightarrow \int \sqrt{a^{2}-x^{2}} d x=\frac{\pi}{2} \sqrt{a^{2}-x^{2}} \frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+C\)
Therefore,
\(\Rightarrow \int \sqrt{4-x^{2}} d x=\frac{\pi}{2} \sqrt{4-x^{2}} \frac{4}{2} \sin ^{-1} \frac{x}{a}+C\)
\(\Rightarrow\frac{x}{2} \sqrt{4-x^{2}}+2 \sin ^{-1} \frac{x}{a}+C\)
2. Integrate: \(\sqrt{1-4 x^{2}}\)
Answer
\(\Rightarrow \sqrt{1-4 x^{2}} d x=\int \sqrt{(1)^{2}-(2 x)^{2} d x}\)
Let \( 2=\mathrm{t} \)
\(\Rightarrow 2 \mathrm{d}x=\mathrm{dt}\)
\(\Rightarrow \mathrm{I}=\frac{1}{2} \int \sqrt{(1)^{2}-(2 x)^{2} d t}\)
We know that,
\(\Rightarrow \int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}} \frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+C\)
Therefore,
\(\Rightarrow \mathrm{I}=\frac{1}{2}\left[\frac{t}{2} \sqrt{1-t^{2}}+\frac{1}{2} \sin ^{-1} t\right]+C\)
\(\Rightarrow\frac{t}{4} \sqrt{1-t^{2}}+\frac{1}{4} \sin ^{-1} t+C\)
\(\Rightarrow\frac{2 x}{4} \sqrt{1-4 x^{2}}+\frac{1}{4} \sin ^{-1} 2 x+C\)
\(\Rightarrow\frac{x}{2} \sqrt{1-4 x^{2}}+\frac{1}{4} \sin ^{-1} 2 x+C\)
exercise 7.7 class 12 maths ncert solutions || class 12 maths ncert solutions chapter 7 exercise 7.7 || integrals class 12 ncert solutions || class 12 maths exercise 7.7 || ex 7.7 class 12 maths ncert solutions
3. Integrate: \(\sqrt{x^{2}+4 x+6}\)
Answer
\(\Rightarrow \mathrm{I}=\int \sqrt{x^{2}+4 x+6} d x\)
\(\Rightarrow \int \sqrt{x^{2}+4 x+4+2} d x\)
\(\Rightarrow \int \sqrt{(x+2)^{2}+(\sqrt{2})^{2}} d x\)
\( \Rightarrow \) We know that,
\(\Rightarrow \int \sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}+\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}+a^{2}}\right|+C}\)
Therefore,
\(\Rightarrow \mathrm{I}=\frac{(x+2)}{2} \sqrt{x^{2}+4 x+6}+\frac{2}{2} \log \left|(x+2)+\sqrt{x^{2}+4 x+6}\right|+C\)
\(\Rightarrow\frac{(x+2)}{2} \sqrt{x^{2}+4 x+6}+\log \left|(x+2)+\sqrt{x^{2}+4 x+6}\right|+C\)
4. Integrate: \(\sqrt{x^{2}+4 x+1}\)
Answer
\(\Rightarrow \mathrm{I}=\int \sqrt{x^{2}+4 x+1} d x\)
\(\Rightarrow\int \sqrt{\left(x^{2}+4 x+4\right)-3} d x\)
\(\Rightarrow\int \sqrt{(x+2)^{2}+(\sqrt{3})^{2}} d x\)
We know that,
\(\int \sqrt{(x+2)^{2}+(\sqrt{3})^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}-\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}+a^{2}}\right|+C\)
Therefore,
\(\Rightarrow \mathrm{I}=\frac{(x+2)}{2} \sqrt{x^{2}+4 x+1}+\frac{3}{2} \log \left|(x+2)+\sqrt{x^{2}+4 x+1}\right|+C\)
5. Integrate: \(\sqrt{1-4 x-x^{2}}\)
Answer
\(\Rightarrow \mathrm{I}=\int \sqrt{1-4 x-x^{2}} d x\)
\(\Rightarrow\int \sqrt{1-\left(x^{2}+4 x+4-4\right)} d x\)
\(\Rightarrow\int \sqrt{1+4-(x+2)^{2}} d x\)
\(\Rightarrow\int \sqrt{(5)^{2}-(x+2)^{2}} d x\)
We know that,
\(\Rightarrow\int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+C\)
\(\Rightarrow \mathrm{I}=\frac{(x+2)}{2} \sqrt{1-4 x-x^{2}}+\frac{5}{2} \sin ^{-1}\left(\frac{x+2}{\sqrt{5}}\right)+C\)
6. Integrate: \(\sqrt{x^{2}+4 x-5}\)
Answer
\(\Rightarrow \mathrm{I}=\int \sqrt{x^{2}+4 x-5} d x\)
\(\Rightarrow\int \sqrt{\left(x^{2}+4 x+4\right)-9} d x\)
\(\Rightarrow\int \sqrt{(x+2)^{2}-(3)^{2}} d x\)
\( \Rightarrow \) We know that,
\(\Rightarrow \int \sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}-\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}+a^{2}}\right|+C\)
Therefore,
\(\Rightarrow \mathrm{I}=\frac{(x+2)}{2} \sqrt{x^{2}+4 x-5}-\frac{9}{2} \log \left|(x+2)+\sqrt{x^{2}+4 x-5}\right|+C\)
exercise 7.7 class 12 maths ncert solutions || class 12 maths ncert solutions chapter 7 exercise 7.7 || integrals class 12 ncert solutions || class 12 maths exercise 7.7 || ex 7.7 class 12 maths ncert solutions
7. Integrate: \(\sqrt{1+3 x-x^{2}}\)
Answer
\(\Rightarrow \mathrm{I}=\int \sqrt{1+3 x-x^{2}} d x\)
\(\Rightarrow\int \sqrt{1-\left(x^{2}-3 x+\frac{9}{4}-\frac{9}{4}\right)} d x\)
\(\Rightarrow\int \sqrt{\left(1+\frac{9}{4}\right)-\left(x+\frac{3}{2}\right)^{2}} d x\)
\(\Rightarrow\int \sqrt{\left(\frac{\sqrt{13}}{2}\right)^{2}-\left(x+\frac{3}{2}\right)^{2}} d x\)
We know that,
\(\Rightarrow\int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+C\)
Therefore,
\(\Rightarrow \mathrm{I}=\frac{x-\frac{3}{2}}{2} \sqrt{1+3 x-x^{2}}+\frac{13}{4 \times 2} \sin ^{-1}\left(\frac{x+\frac{3}{2}}{\frac{\sqrt{13}}{2}}\right)+C\)
\(\Rightarrow\frac{2 x-3}{4} \sqrt{1+3 x-x^{2}}+\frac{13}{8} \sin ^{-1}\left(\frac{2 x-3}{\sqrt{13}}\right)+C\)
8. Integrate: \(\sqrt{x^{2}+3 x}\)
Answer
\(\Rightarrow \mathrm{I}=\int \sqrt{x^{2}+3 x} d x\)
\(\Rightarrow\int \sqrt{x^{2}-3 x+\frac{9}{4}-\frac{9}{4}} d x\)
\(\Rightarrow\int \sqrt{\left(x+\frac{3}{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}} d x\)
We know that,
\(\Rightarrow\int \sqrt{x^{2}-a^{2} x} d x=\frac{x}{2} \sqrt{x^{2}-a^{2}}+\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}-a^{2}}\right|+C\)
Therefore,
\(\Rightarrow \mathrm{I}=\frac{\left(x+\frac{3}{2}\right)}{2} \sqrt{x^{2}+3 x}-\frac{\frac{9}{4}}{2} \log \left|\left(x+\frac{3}{2}\right)+\sqrt{x^{2}-3 x}\right|+C\)
\(\Rightarrow\frac{(2 x+3)}{4} \sqrt{x^{2}+3 x}-\frac{9}{8} \log \left|\left(x+\frac{3}{2}\right)+\sqrt{x^{2}-3 x}\right|+C\)
9. Integrate: \(\sqrt{1+\frac{x^{2}}{9}}\)
Answer
\(\Rightarrow \mathrm{I}=\int \sqrt{1+\frac{x^{2}}{9}} d x\)
\(\Rightarrow\frac{1}{3} \int \sqrt{9+x^{2}} d x\)
\(\Rightarrow\frac{1}{3} \int \sqrt{(3)^{2}-x^{2}} d x\)
We know that,
\(\Rightarrow\int \sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \log \left|x^{2}+a^{2}\right|+C\)
Therefore,
\(\Rightarrow \mathrm{I}=\frac{1}{3}\left[\frac{x}{2} \sqrt{x^{2}+9}+\frac{9}{2} \log \left|x+\sqrt{x^{2}+9}\right|\right]+C\)
\(\Rightarrow\frac{x}{6} \sqrt{x^{2}+9}+\frac{3}{2} \log \left|x+\sqrt{x^{2}+9}\right|+C\)
10. Choose the correct answer:
\( \int \sqrt{1+x^{2}} d x \) is equal to
A. \( \frac{x}{2} \sqrt{1+x^{2}+\frac{1}{2}} \log \left|x+\sqrt{1+x^{2}}\right|+C \)
B. \( \frac{2}{3}\left(1+x^{2}\right)^{\frac{3}{2}}+C \)
C. \( \frac{2}{3} x\left(1+x^{2}\right)^{\frac{3}{2}}+C \)
D. \( \frac{x}{2} \sqrt{1+x^{2}}+\frac{1}{2} x^{2} \log \left|\left(x+\sqrt{1+x^{2}}\right)\right|+C \)
\( \int \sqrt{1+x^{2}} d x \) is equal to
A. \( \frac{x}{2} \sqrt{1+x^{2}+\frac{1}{2}} \log \left|x+\sqrt{1+x^{2}}\right|+C \)
B. \( \frac{2}{3}\left(1+x^{2}\right)^{\frac{3}{2}}+C \)
C. \( \frac{2}{3} x\left(1+x^{2}\right)^{\frac{3}{2}}+C \)
D. \( \frac{x}{2} \sqrt{1+x^{2}}+\frac{1}{2} x^{2} \log \left|\left(x+\sqrt{1+x^{2}}\right)\right|+C \)
Answer
We know that,
\(\Rightarrow\int \sqrt{a^{2}+x^{2}} d x=\frac{x}{2} \sqrt{a^{2}+x^{2}}+\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}+a^{2}}\right|+C\)
Therefore,
\(\Rightarrow \int \sqrt{1+x^{2}} d x=\frac{x}{2} \sqrt{1+x^{2}}+\frac{1}{2} \log \left|x+\sqrt{1+x^{2}}\right|+C\)
11. Choose the correct answer:
\( \int \sqrt{x^{2}-8 x+7} d x \) is equal to
A. \( \frac{1}{2}(x-4) \sqrt{x^{2}-2 x+7}+9 \log \left|x-4+\sqrt{x^{2}-8 x+7}\right|+C \)
B. \( \frac{1}{2}(x+4) \sqrt{x^{2}-8 x+7}+9 \log \left|x+4+\sqrt{x^{2}-8 x+7}\right|+C \)
C. \( \frac{1}{2}(x-4) \sqrt{x^{2}-8 x+7}-3 \sqrt{2} \log \left|x-4+\sqrt{x^{2}-8 x+7}\right|+C \)
D. \( \frac{1}{2}(x-4) \sqrt{x^{2}-8 x+7}-\frac{9}{2} \log \left|x-4+\sqrt{x^{2}-8 x+7}\right|+C \)
\( \int \sqrt{x^{2}-8 x+7} d x \) is equal to
A. \( \frac{1}{2}(x-4) \sqrt{x^{2}-2 x+7}+9 \log \left|x-4+\sqrt{x^{2}-8 x+7}\right|+C \)
B. \( \frac{1}{2}(x+4) \sqrt{x^{2}-8 x+7}+9 \log \left|x+4+\sqrt{x^{2}-8 x+7}\right|+C \)
C. \( \frac{1}{2}(x-4) \sqrt{x^{2}-8 x+7}-3 \sqrt{2} \log \left|x-4+\sqrt{x^{2}-8 x+7}\right|+C \)
D. \( \frac{1}{2}(x-4) \sqrt{x^{2}-8 x+7}-\frac{9}{2} \log \left|x-4+\sqrt{x^{2}-8 x+7}\right|+C \)
Answer
\(\Rightarrow \mathrm{I}=\int \sqrt{x^{2}-8 x+7} d x\)
\(\Rightarrow\int \sqrt{\left(x^{2}-8 x+16\right)-9} d x\)
\(\Rightarrow\int \sqrt{(x-4)^{2}-(3)^{2}} d x\)
\( \Rightarrow \) We know that,
\(\Rightarrow \int \sqrt{x^{2}-a^{2}} d x=\frac{x}{2} \sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}-a^{2}}\right|+C\)
Therefore,
\(\Rightarrow \mathrm{I}=\frac{(x-4)}{2} \sqrt{x^{2}-8 x+7}-\frac{9}{2} \log \left|(x+4)+\sqrt{x^{2}-8 x+7}\right|+C\)
