Ex 7.8 class 12 maths ncert solutions | class 12 maths exercise 7.8 | class 12 maths ncert solutions chapter 7 exercise 7.8 | exercise 7.8 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.
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Exercise 7.8
1. Evaluate using limit of sums \( \int_{a}^{b} x\mathrm{d}x \)
Answer
\( f(x) \) is continuous in \( [a, b] \)
\( \int_{a}^{b} \mathrm{f}(x) \mathrm{d}x=\lim _{n \rightarrow \infty} h \sum_{r=0}^{n-1} \mathrm{f}(\mathrm{a}+r h) \), where \( \mathrm{h}=\frac{b-a}{n} \)
here \( \mathrm{h}=\mathrm{b}-\frac{ \mathrm{a} }{ \mathrm{n} } \)
\(\int_{a}^{b}(x) \mathrm{d}x=\lim _{n \rightarrow \infty}\left(\frac{b-a}{n}\right) \sum_{r=0}^{n-1} \mathrm{f}\left(\mathrm{a}+\frac{(b-a) r}{n}\right)\)
\(=\lim _{n \rightarrow \infty}\left(\frac{b-a}{n}\right) \sum_{r=0}^{n-1} \mathrm{f}\left(\frac{(b-a) r}{n}\right)+\mathrm{a}\)
\(=\lim _{n \rightarrow \infty}\left(\frac{b-a}{n}\right)\left(\frac{(b-a)(n-1)(n)}{2 n}+a(n-1)\right)\)
\(=\lim _{n \rightarrow \infty}\left(\frac{b-a}{n}\right) \cdot \frac{(b-a)\left(n^{2}-n\right)+2 a n^{2}-2 a n}{2 n}\)
\(=\lim _{n \rightarrow \infty}\left(\frac{b-a}{n}\right) \cdot \frac{(b+a) n^{2}-(b+a) n}{2 n}\)
\(=\lim _{n \rightarrow \infty} \frac{(b+a)(b-a) n^{2}-(b+a)(b-a) n}{2 n^{2}}\)
\(=\lim _{n \rightarrow \infty}\left(\frac{(b+a)(b-a)}{2}-\frac{(b+a)(b-a)}{n}\right)\)
\(=\frac{(b+a)(b-a)}{2}\)
\(=\frac{b^{2}-a^{2}}{2}\)
2. Evaluate using limit of sums \( \int_{0}^{5}(x+1) d x \)
Answer
\( \mathrm{f}(x) \) is continuous in \( [0,5] \)
\( \int_{a}^{b} \mathrm{f}(x) \mathrm{d}x=\lim _{n \rightarrow \infty} h \sum_{r=0}^{n-1} \mathrm{f}(\mathrm{a}+\mathrm{rh}) \), where \( \mathrm{h}=\frac{b-a}{n} \)
here \( \mathrm{h=}\frac{ 5 }{ \mathrm{n} } \)
\(\int_{0}^{5}(x+1) \mathrm{d}x=\lim _{n \rightarrow \infty}\left(\frac{5}{n}\right) \sum_{r=0}^{n-1} \mathrm{f}\left(\frac{5 \mathrm{r}}{\mathrm{n}}\right)\)
\(=\lim _{n \rightarrow \infty}\left(\frac{5}{n}\right) \sum_{r=0}^{n-1} \mathrm{f}\left(\frac{5 \mathrm{r}}{\mathrm{n}}\right)+1\)
\(=\lim _{n \rightarrow \infty}\left(\frac{5}{n}\right)\left(\frac{5(n-1)(n)}{2 n}+(n-1)\right)\)
\(=\lim _{n \rightarrow \infty} \frac{5}{n} \cdot \frac{5 n^{2}-5 n+2 n^{2}-2 n}{2 n}\)
\(=\lim _{n \rightarrow \infty} \frac{5}{n} \cdot \frac{7 n^{2}-7 n}{2 n}\)
\(=\lim _{n \rightarrow \infty} \frac{35 n^{2}-35 n}{2 n^{2}}\)
\(=\lim _{n \rightarrow \infty} \frac{35}{2}-\left(\frac{35}{2 n}\right)\)
\(=\frac{35}{2}\)
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3. Evaluate using limit of sums \( \int_{2}^{3} x^{2} \mathrm{d}x \)
Answer
\( f(x) \) is continuous in [2,3]
\( \int_{a}^{b} \mathrm{f}(x) \mathrm{d}x=\lim _{n \rightarrow \infty} h \sum_{r=0}^{n-1} \mathrm{f}(\mathrm{a}+\mathrm{rh}) \), where \( \mathrm{h}=\frac{b-a}{n} \)
here \(\mathrm{h=} \frac{ 1 }{ \mathrm{n} } \)
\(\int_{2}^{3} x^{2} \mathrm{d}x=\lim _{n \rightarrow \infty}\left(\frac{1}{n}\right) \sum_{r=0}^{n-1} \mathrm{f}\left(2+\left(\frac{\mathrm{r}}{\mathrm{n}}\right)\right)\)
\(=\lim _{n \rightarrow \infty}\left(\frac{1}{n}\right) \sum_{r=0}^{n-1}\left(2+\left(\frac{\mathrm{r}}{\mathrm{n}}\right)\right)^{2}\)
\(=\lim _{n \rightarrow \infty}\left(\frac{1}{n}\right) \sum_{r=0}^{n-1}\left(\frac{r^{2}}{n^{2}}+4+\frac{4 r}{n}\right)\)
\(=\lim _{n \rightarrow \infty} \frac{1}{n}\left(\frac{(n-1)(n)(2 n-1)}{6 n^{2}}+4 n+\frac{4(n-1)(n)}{2 n}\right)\)
\(=\lim _{n \rightarrow \infty} \frac{1}{n}\left(\frac{\left(n^{2}-n\right)(2 n-1)}{6 n^{2}}+4 n+\frac{2\left(n^{2}-n\right)}{n}\right)\)
\(=\lim _{n \rightarrow \infty} \frac{1}{n}\left(\frac{\left(2 n^{3}-2 n^{2}-n^{2}+n\right)}{6 n^{2}}+4 n+\frac{2\left(n^{2}-n\right)}{n}\right)\)
\(=\lim _{n \rightarrow \infty} \frac{1}{n}\left(\frac{\left(2 n^{3}-3 n^{2}+n\right)+\left(24 n^{3}\right)+\left(12 n^{3}-12 n^{2}\right)}{6 n^{2}}\right)\)
\(=\lim _{n \rightarrow \infty} \frac{1}{n}\left(\frac{38 n^{3}-15 n^{2}+n}{6 n^{2}}\right)\)
\(=\lim _{n \rightarrow \infty}\left(\frac{38 n^{3}-15 n^{2}+n}{6 n^{2}}\right)\)
\(=\lim _{n \rightarrow \infty}\left(\frac{38}{6}\right)-\left(\frac{15}{6 n}\right)+\left(\frac{1}{6 n^{2}}\right)\)
\(=\frac{38}{6}\)
\(=\frac{19}{3}\)
4. Evaluate using limit of sums \( \int_{1}^{4}\left(x^{2}-x\right) d x \)
Answer
\( \mathrm{f}(x) \) is continuous in [1,4]
\( \int_{a}^{b} \mathrm{f}(x) \mathrm{d}x=\lim _{n \rightarrow \infty} h \sum_{r=0}^{n-1} \mathrm{f}(\mathrm{a}+\mathrm{rh}) \), where \( \mathrm{h}=\frac{b-a}{n} \)
here \(\mathrm{h=} \frac{ 3 }{ \mathrm{n} } \)
\(\int_{1}^{4}\left(x^{2}-x\right) \mathrm{d}x=\lim _{n \rightarrow \infty}\left(\frac{3}{n}\right) \sum_{r=0}^{n-1} \mathrm{f}\left(\left(1+\frac{3 \mathrm{r}}{\mathrm{n}}\right)\right)\)
\(=\lim _{n \rightarrow \infty}\left(\frac{3}{n}\right) \sum_{r=0}^{n-1}\left(\left(1+\frac{3 \mathrm{r}}{\mathrm{n}}\right)^{2}-\left(1+\frac{3 \mathrm{r}}{\mathrm{n}}\right)\right)\)
\(=\lim _{n \rightarrow \infty}\left(\frac{3}{n}\right) \sum_{r=0}^{n-1}\left(1+\frac{9 r^{2}}{n^{2}}+\frac{6 r}{n}-1-\frac{3 r}{n}\right)\)
\(=\lim _{n \rightarrow \infty}\left(\frac{3}{n}\right) \sum_{r=0}^{n-1}\left(\frac{9 r^{2}}{n^{2}}+\frac{3 r}{n}\right)\)
\(=\lim _{n \rightarrow \infty} \frac{3}{n}\left(\frac{9(n-1)(n)(2 n-1)}{6 n^{2}}+\frac{3 n(n-1)}{2 n}\right)\)
\(=\lim _{n \rightarrow \infty} \frac{3}{n}\left(\frac{9\left(n^{2}-n\right)(2 n-1)}{6 n^{2}}+\frac{3 n(n-1)}{2 n}\right)\)
\(=\lim _{n \rightarrow \infty} \frac{3}{n}\left(\frac{9\left(2 n^{3}-2 n^{2}-n^{2}+n\right)}{6 n^{2}}+\frac{3 n(n-1)}{2 n}\right)\)
\(=\lim _{n \rightarrow \infty} \frac{3}{n}\left(\frac{\left(18 n^{3}-27 n^{2}+9 n\right)+\left(9 n^{3}-9 n^{2}\right)}{6 n^{2}}\right)\)
\(=\lim _{n \rightarrow \infty} \frac{3}{n}\left(\frac{27 n^{3}-36 n^{2}+9 n}{6 n^{2}}\right)\)
\(=\lim _{n \rightarrow \infty}\left(\frac{81 n^{3}-108 n^{2}+27 n}{6 n^{3}}\right)\)
\(=\lim _{n \rightarrow \infty}\left(\frac{81}{6}\right)-\left(\frac{108}{6 n}\right)+\left(\frac{27}{6 n^{2}}\right)\)
\(=\frac{27}{2}\)
5. Evaluate using limit of sums \( \int_{-1}^{1} \mathrm{e}^{x} \mathrm{d}x \)
Answer
\( \mathrm{f}(x) \) is continuous in \( [1,4] \)
\( \int_{a}^{b} \mathrm{f}(x) \mathrm{d}x=\lim _{n \rightarrow \infty} h \sum_{r=0}^{n-1} \mathrm{f}(\mathrm{a}+\mathrm{rh}) \), where \( \mathrm{h}=\frac{b-a}{n} \)
here \(\mathrm{h=}\frac{ 2 }{ \mathrm{n} } \)
\(=\int_{0}^{2}\left(\mathrm{e}^{x}\right) \mathrm{d}x=\lim _{n \rightarrow \infty}\left(\frac{2}{n}\right) \sum_{r=0}^{n-1} \mathrm{f}\left(-1+\frac{2 \mathrm{r}}{\mathrm{n}}\right)\)
\(=\lim _{n \rightarrow \infty}\left(\frac{2}{n}\right) \sum_{r=0}^{n-1} \mathrm{e}^{\frac{2 r}{n}-1}\)
\(=\lim _{n \rightarrow \infty}\left(\frac{2}{n}\right)\left(e^{0}+e^{h}+e^{2 h}+\ldots \ldots \ldots .+e^{n h}\right)\)
\(\text {Sum of }=e^{0}+e^{h}+e^{2 h}+\ldots \ldots \ldots .+e^{n h}\)
Which is g.p with common ratio \( \frac{ \mathrm{e^{1}} }{ \mathrm{n} } \).
\(\text { Whose sum is }=\frac{e^{h}\left(1-e^{n h}\right)}{1-e^{h}}\)
\(=\lim _{n \rightarrow \infty}\left(\frac{2}{n e}\right)\left(\frac{e^{h}\left(1-e^{n h}\right)}{1-e^{h}}\right)\)
\(=\lim _{n \rightarrow \infty}\left(\frac{2}{n e}\right) \cdot \frac{e^{h}\left(1-e^{n h}\right)}{\frac{1-e^{h} \cdot h}{h}}\)
\(=-1\)
\(=\lim _{n \rightarrow \infty}\left(\frac{2}{n e}\right) \cdot\left(\frac{e^{h}\left(1-e^{n h}\right)}{h}\right)\)
As \( \frac{ \mathrm{h}=2 }{ \mathrm{n} } \)
\(=\lim _{n \rightarrow \infty}\left(\frac{2}{n e}\right) \cdot\left(\frac{e^{\left(\frac{2}{n}\right)\left(1-e^{n \times\left(\frac{2}{n}\right)}\right)}}{-\frac{2}{n}}\right)\)
\(=\frac{e^{2}-1}{e}\)
\(=\mathrm{e}-\mathrm{e}^{-1}\)
6. Evaluate using limit of sums \( \int_{0}^{4}\left(x+\mathrm{e}^{2 x}\right) \mathrm{d} x\)
Answer
\(h(x)=\int_{0}^{4} x \cdot d x\)
\(g(x)=\int_{0}^{4} e^{2 x} \cdot d x\)
\(F(x)=h(x)+g(x)\)
Solving for \( \mathrm{h}(x) \)
\( \mathrm{h}(x) \) is continuous in [0,4]
\( \int_{a}^{b} \mathrm{f}(x) \mathrm{d}x=\lim _{n \rightarrow \infty} h \sum_{r=0}^{n-1} \mathrm{f}(\mathrm{a}+\mathrm{rh}) \), where \( \mathrm{h}=\frac{b-a}{n} \)
here \( h=\frac{ 4 }{ n } \)
\(\int_{0}^{4}(x) \mathrm{dx}=\lim _{n \rightarrow \infty}\left(\frac{4}{n}\right) \sum_{r=0}^{n-1} \mathrm{f}\left(\frac{4 \mathrm{r}}{\mathrm{n}}\right)\)
\(=\lim _{n \rightarrow \infty}\left(\frac{4}{n}\right) \sum_{r=0}^{n-1}\left(\frac{4 \mathrm{r}}{\mathrm{n}}\right)\)
\(=\lim _{n \rightarrow \infty}\left(\frac{4}{n}\right)\left(\frac{2(n-1)(n)}{n}\right)\)
\(=\lim _{n \rightarrow \infty}\left(\frac{4}{n}\right) \cdot \frac{2 n^{2}-2 n}{n}\)
\(=\lim _{n \rightarrow \infty} \frac{4}{n} \cdot\frac{2 n^{2}-2 n}{n}\)
\(=\lim _{n \rightarrow \infty} \frac{8 n^{2}-8 n}{n^{2}}\)
\(=\lim _{n \rightarrow \infty} 8-\left(\frac{8}{n}\right)\)
\(=8\)
Now solving for \( \mathrm{g}(x) \)
\( \mathrm{g}(x) \) is continuous in \( [0,4] \)
\( \int_{a}^{b} \mathrm{f}(x) \mathrm{d}x=\lim _{n \rightarrow \infty} h \sum_{r=0}^{n-1} \mathrm{f}(\mathrm{a}+\mathrm{rh}) \), where \( \mathrm{h}=\frac{b-a}{n} \)
here \(\mathrm{h=}\frac{ 4 }{ \mathrm{n} } \)
\(=\int_{0}^{4}\left(\mathrm{e}^{2 x}\right) \mathrm{d}x=\lim _{n \rightarrow \infty}\left(\frac{4}{n}\right) \sum_{r=0}^{n-1} \mathrm{f}\left(\frac{4 \mathrm{r}}{\mathrm{n}}\right)\)
\(=\lim _{n \rightarrow \infty}\left(\frac{4}{n}\right) \sum_{r=0}^{n-1} \mathrm{e}^{\frac{4 r}{n}}\)
\(=\lim _{n \rightarrow \infty}\left(\frac{4}{n}\right)\left(e^{0}+e^{h}+e^{2 h}+ \ldots \ldots \ldots \ldots+e^{n h}\right)\)
\(\text {Sum of }=e^{0}+e^{h}+e^{2 h}+ \ldots \ldots \ldots .+e^{n h}\)
Which is g.p with common ratio \( e^{\frac{ 1 }{ n }} \)
\(\text { Whose sum is }=\frac{e^{h}\left(1-e^{n h}\right)}{1-e^{h}}\)
\(=\lim _{n \rightarrow \infty}\left(\frac{4}{n e}\right)\left(\frac{e^{h}\left(1-e^{n h}\right)}{1-e^{h}}\right)\)
\(=\lim _{n \rightarrow \infty}\left(\frac{4}{n e}\right) \cdot \frac{e^{h}\left(1-e^{n h}\right)}{\frac{1-e^{h} \cdot h}{h}}\)
\(=\lim _{n \rightarrow \infty} \frac{1-e^{h}}{h}=-1\)
\(=\lim _{n \rightarrow \infty}\left(\frac{4}{n}\right)\left(\frac{e^{h}\left(1-e^{n h}\right)}{-h}\right)\)
As \( \mathrm{h}=\frac{ 4 }{ \mathrm{n} } \)
\(=\lim _{n \rightarrow \infty}\left(\frac{4}{n}\right) \cdot\left(\frac{e^{\left(\frac{4}{n}\right)\left(1-e^{n \times\left(\frac{4}{n}\right)}\right)}}{-\frac{4}{n}}\right)\)
\(=\left(\mathrm{e}^{8}-1\right)\)
Now for \( \mathrm{f}(x)=\mathrm{h}(x)+\mathrm{g}(x) \)
\(=8+e^{8}-1\)
