SATHEE: Indefinite Integration 3 Question 13

Indefinite Integration 3 Question 13

13. The value of the definite integral $\int_{0}^{1} (1 + e^{- x^{2}}) d x$ is

(a) -1

(b) 2

(1981, 2M)

(d) None of the above

Objective Question II

(One or more than one correct option)

Show Answer

Answer:

Correct Answer: 13. (d)

Solution:

If $f (x)$ is a continuous function defined on $[a, b]$ , then

$m (b - a) \leq \int_{a}^{b} f (x) d x \leq M (b - a)$

where, $M$ and $m$ are maximum and minimum values respectively of $f (x)$ in $[a, b]$ .

Here, $f (x) = 1 + e^{- x^{2}}$ is continuous in $[0, 1]$ .

Now, $0 < x < 1 \Rightarrow x^{2} < x \Rightarrow e^{x^{2}} < e^{x} \Rightarrow e^{- x^{2}} > e^{- x}$

Again, $0 < x < 1 \Rightarrow x^{2} > 0 \Rightarrow e^{x^{2}} > e^{0} \Rightarrow e^{- x^{2}} < 1$

$\begin{aligned} ∴ e^{- x} < e^{- x^{2}} < 1, \forall x \in [0, 1] \\ \Rightarrow 1 + e^{- x} < 1 + e^{- x^{2}} < 2, \forall x \in [0, 1] \\ \Rightarrow \int_{0}^{1} (1 + e^{- x}) d x < \int_{0}^{1} (1 + e^{- x^{2}}) d x < \int_{0}^{1} 2 d x \\ \Rightarrow 2 - \frac{1}{e} < \int_{0}^{1} (1 + e^{- x^{2}}) d x < 2 \end{aligned}$

Indefinite Integration 3 Question 13

13. The value of the definite integral $\int_{0}^{1} (1 + e^{- x^{2}}) d x$ is

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Indefinite Integration 3 Question 13

13. The value of the definite integral ∫01(1+e−x2)dx is

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu

13. The value of the definite integral $\int_{0}^{1} (1 + e^{- x^{2}}) d x$ is