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Integrals Class 12 Solutions: Exercise 7.8

Topic: Definite Integrals as a Limit of a Sum

Question 1: Evaluate $\displaystyle \int_a^b x \, dx$ using limit of sums.

Solution:

We know that $\displaystyle \int_a^b f(x) \, dx = \lim_{h \to 0} h \sum_{r=0}^{n-1} f(a+rh)$, where $h = \frac{b-a}{n}$.

Here, $f(x) = x$, $a = a$, $b = b$, and $h = \frac{b-a}{n}$.
$f(a+rh) = a + rh$

Substituting into the formula:

$$ \begin{aligned} I &= \lim_{n \to \infty} h \sum_{r=0}^{n-1} (a + rh) \\ &= \lim_{n \to \infty} h \left[ \sum_{r=0}^{n-1} a + h \sum_{r=0}^{n-1} r \right] \\ &= \lim_{n \to \infty} h \left[ na + h \frac{(n-1)n}{2} \right] \end{aligned} $$

Since $nh = b-a$, we substitute $h = \frac{b-a}{n}$:

$$ \begin{aligned} I &= \lim_{n \to \infty} \left[ (nh)a + \frac{(nh)(nh - h)}{2} \right] \\ &= (b-a)a + \frac{(b-a)(b-a - 0)}{2} \quad (\text{as } h \to 0) \\ &= ab - a^2 + \frac{(b-a)^2}{2} \\ &= \frac{2ab - 2a^2 + b^2 + a^2 - 2ab}{2} \end{aligned} $$

Answer: $\displaystyle \frac{b^2 - a^2}{2}$

Question 2: Evaluate $\displaystyle \int_0^5 (x+1) \, dx$ using limit of sums.

Solution:

Let $I = \int_0^5 (x+1) \, dx$. Here $a=0, b=5$, so $h = \frac{5-0}{n} = \frac{5}{n}$ or $nh=5$.

$f(x) = x+1$
$f(a+rh) = f(0+rh) = f(rh) = rh + 1$

Using the limit of sum formula:

$$ \begin{aligned} I &= \lim_{n \to \infty} h \sum_{r=0}^{n-1} (rh + 1) \\ &= \lim_{n \to \infty} h \left[ h \sum_{r=0}^{n-1} r + \sum_{r=0}^{n-1} 1 \right] \\ &= \lim_{n \to \infty} h \left[ h \frac{n(n-1)}{2} + n \right] \end{aligned} $$

Substitute $h = 5/n$:

$$ \begin{aligned} I &= \lim_{n \to \infty} \frac{5}{n} \left[ \frac{5}{n} \frac{n(n-1)}{2} + n \right] \\ &= \lim_{n \to \infty} \left[ \frac{25}{2} \left(1 - \frac{1}{n}\right) + 5 \right] \\ &= \frac{25}{2}(1 - 0) + 5 \\ &= \frac{25}{2} + \frac{10}{2} \end{aligned} $$

Answer: $\displaystyle \frac{35}{2}$

Question 3: Evaluate $\displaystyle \int_2^3 x^2 \, dx$ using limit of sums.

Solution:

Here $a=2, b=3$, so $h = \frac{3-2}{n} = \frac{1}{n}$ and $nh=1$.

$f(x) = x^2$
$f(a+rh) = f(2+rh) = (2+rh)^2 = 4 + 4rh + r^2h^2$

$$ \begin{aligned} I &= \lim_{n \to \infty} h \sum_{r=0}^{n-1} (4 + 4rh + r^2h^2) \\ &= \lim_{n \to \infty} h \left[ 4n + 4h \frac{n(n-1)}{2} + h^2 \frac{n(n-1)(2n-1)}{6} \right] \end{aligned} $$

Multiply $h$ inside and substitute $nh=1$:

$$ \begin{aligned} I &= \lim_{n \to \infty} \left[ 4nh + 2(nh)(nh-h) + \frac{(nh)(nh-h)(2nh-h)}{6} \right] \\ &= \lim_{n \to \infty} \left[ 4(1) + 2(1)(1-h) + \frac{1(1-h)(2-h)}{6} \right] \end{aligned} $$

As $n \to \infty, h \to 0$:

$$ \begin{aligned} I &= 4 + 2(1) + \frac{1(1)(2)}{6} \\ &= 4 + 2 + \frac{1}{3} = 6 + \frac{1}{3} \end{aligned} $$

Answer: $\displaystyle \frac{19}{3}$

Question 4: Evaluate $\displaystyle \int_1^4 (x^2 - x) \, dx$ using limit of sums.

Solution:

Here $a=1, b=4$, so $h = \frac{3}{n}$, $nh=3$.

$f(x) = x^2 - x$
$f(1+rh) = (1+rh)^2 - (1+rh) = 1 + 2rh + r^2h^2 - 1 - rh = rh + r^2h^2$

$$ \begin{aligned} I &= \lim_{n \to \infty} h \sum_{r=0}^{n-1} (rh + r^2h^2) \\ &= \lim_{n \to \infty} h \left[ h \frac{n(n-1)}{2} + h^2 \frac{n(n-1)(2n-1)}{6} \right] \\ &= \lim_{n \to \infty} \left[ \frac{(nh)(nh-h)}{2} + \frac{(nh)(nh-h)(2nh-h)}{6} \right] \end{aligned} $$

Substitute $nh=3$ and let $h \to 0$:

$$ \begin{aligned} I &= \frac{3(3-0)}{2} + \frac{3(3-0)(6-0)}{6} \\ &= \frac{9}{2} + \frac{54}{6} \\ &= 4.5 + 9 \end{aligned} $$

Answer: $\displaystyle \frac{27}{2}$

Question 5: Evaluate $\displaystyle \int_{-1}^1 e^x \, dx$ using limit of sums.

Solution:

Here $a=-1, b=1$, so $h = \frac{2}{n}$, $nh=2$.

$f(x) = e^x$
$f(a+rh) = e^{-1+rh} = e^{-1} \cdot e^{rh}$

This forms a Geometric Progression (G.P.).

$$ \begin{aligned} I &= \lim_{n \to \infty} h \sum_{r=0}^{n-1} e^{-1} \cdot e^{rh} \\ &= \lim_{n \to \infty} h e^{-1} \left( 1 + e^h + e^{2h} + \dots + e^{(n-1)h} \right) \end{aligned} $$

Using sum of G.P. formula $S_n = \frac{a(R^n-1)}{R-1}$ where $R=e^h$:

$$ \begin{aligned} I &= \lim_{h \to 0} h e^{-1} \frac{1 \cdot (e^{nh} - 1)}{e^h - 1} \\ &= e^{-1}(e^2 - 1) \lim_{h \to 0} \frac{h}{e^h - 1} \end{aligned} $$

Since $\lim_{h \to 0} \frac{h}{e^h - 1} = 1$:

$$ I = e^{-1}(e^2 - 1) = e - e^{-1} $$

Answer: $\displaystyle e - \frac{1}{e}$

Question 6: Evaluate $\displaystyle \int_0^4 (x + e^{2x}) \, dx$ using limit of sums.

Solution:

We can split this into two integrals: $I = I_1 + I_2$, where $I_1 = \int_0^4 x \, dx$ and $I_2 = \int_0^4 e^{2x} \, dx$.

Part 1: $I_1 = \int_0^4 x \, dx$
Similar to Question 1 with $a=0, b=4$. The result is $\frac{4^2 - 0^2}{2} = 8$.

Part 2: $I_2 = \int_0^4 e^{2x} \, dx$
$a=0, b=4, h=4/n$. $f(rh) = e^{2rh}$.

$$ \begin{aligned} I_2 &= \lim_{n \to \infty} h \sum_{r=0}^{n-1} e^{2rh} \\ &= \lim_{h \to 0} h \left( 1 + e^{2h} + e^{4h} + \dots + e^{2(n-1)h} \right) \end{aligned} $$

This is a G.P. with common ratio $R = e^{2h}$:

$$ \begin{aligned} I_2 &= \lim_{h \to 0} h \frac{(e^{2h})^n - 1}{e^{2h} - 1} \\ &= \lim_{h \to 0} h \frac{e^{2nh} - 1}{e^{2h} - 1} \end{aligned} $$

Substitute $nh = 4$:

$$ \begin{aligned} I_2 &= (e^8 - 1) \lim_{h \to 0} \frac{h}{e^{2h} - 1} \\ &= (e^8 - 1) \lim_{h \to 0} \frac{1}{2} \cdot \frac{2h}{e^{2h} - 1} \\ &= (e^8 - 1) \cdot \frac{1}{2} \cdot 1 \end{aligned} $$

Combining both parts:
$I = I_1 + I_2 = 8 + \frac{e^8 - 1}{2} = \frac{16 + e^8 - 1}{2}$

Answer: $\displaystyle \frac{15 + e^8}{2}$

OMTEX AD 2

Integrals Class 12 Exercise 7.8 Solutions - Limit of Sums

Integrals Class 12 Solutions: Exercise 7.8