Evaluate the definite integral int_{-1}^{1}(x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $I = \int_{-1}^{1} (x+1) d x$. First,find the indefinite integral: $\int (x+1) d x = \frac{x^2}{2} + x = F(x)$. By the second fundamental theo

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) Let $I = \int_{-1}^{1} (x+1) d x$. First,find the indefinite integral: $\int (x+1) d x = \frac{x^2}{2} + x = F(x)$. By the second fundamental theorem of calculus,we have: $I = F(1) - F(-1)$. Substitute the limits: $F(1) = \frac{1^2}{2} + 1 = \frac{1}{2} + 1 = \frac{3}{2}$. $F(-1) = \frac{(-1)^2}{2} + (-1) = \frac{1}{2} - 1 = -\frac{1}{2}$. Calculate the difference: $I = \frac{3}{2} - (-\frac{1}{2}) = \frac{3}{2} + \frac{1}{2} = \frac{4}{2} = 2$.

Evaluate the definite integral $\int_{-1}^{1}(x+1) d x$.

Similar Questions

For $m, n > 0$,let $\alpha(m, n)=\int_0^2 t^m(1+3 t)^n d t$. If $11 \alpha(10,6)+18 \alpha(11,5)= p (14)^6$,then $p$ is equal to $......$.

The value of $\int_0^{2\pi} |\sin^3 \theta| \, d\theta$ is

$\int_{ - \pi /4}^{\pi /2} {{e^{ - x}}\sin x\,dx} = $

Evaluate $\int_{0}^{2}(x^{2}+3) dx$ as a limit of sums.

$\int_{0}^{\infty} \frac{dx}{(x^{2}+4)(x^{2}+9)} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module