int_{0}^{sqrt{3}} (x+4)^2 e^{x^2} dx + int_{s | vedclass

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

Question

(B) We are given the expression $I = \int_{0}^{\sqrt{3}} (x+4)^2 e^{x^2} dx + \int_{\sqrt{3}}^{0} (x-4)^2 e^{x^2} dx$. Using the property $\int_{b}^{a

Vedclass · Accepted Answer

$8(e^3 - 1)$

Vedclass · Answer

(B) We are given the expression $I = \int_{0}^{\sqrt{3}} (x+4)^2 e^{x^2} dx + \int_{\sqrt{3}}^{0} (x-4)^2 e^{x^2} dx$. Using the property $\int_{b}^{a} f(x) dx = -\int_{a}^{b} f(x) dx$,we can rewrite the second integral as $-\int_{0}^{\sqrt{3}} (x-4)^2 e^{x^2} dx$. Thus,$I = \int_{0}^{\sqrt{3}} e^{x^2} \{(x+4)^2 - (x-4)^2\} dx$. Expanding the terms inside the bracket: $(x+4)^2 - (x-4)^2 = (x^2 + 8x + 16) - (x^2 - 8x + 16) = 16x$. So,$I = \int_{0}^{\sqrt{3}} e^{x^2} (16x) dx$. Let $u = x^2$,then $du = 2x dx$,which implies $8 du = 16x dx$. When $x=0, u=0$. When $x=\sqrt{3}, u=3$. $I = \int_{0}^{3} 8 e^u du = 8 [e^u]_{0}^{3} = 8(e^3 - e^0) = 8(e^3 - 1)$.

$\int_{0}^{\sqrt{3}} (x+4)^2 e^{x^2} dx + \int_{\sqrt{3}}^{0} (x-4)^2 e^{x^2} dx$ is equal to

Similar Questions

The value of $\int_0^{\pi /2} \frac{dx}{1 + \tan^3 x}$ is

Show that $\int_{0}^{a} f(x) g(x) \, dx = 2 \int_{0}^{a} f(x) \, dx$,if $f(x) = f(a-x)$ and $g(x) + g(a-x) = 4$.

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \log \left(\frac{2-\sin x}{2+\sin x}\right) d x=$

Let $T > 0$ be a fixed number. $f: R \rightarrow R$ is a continuous function such that $f(x+T) = f(x)$ for all $x \in R$. If $I = \int_0^T f(x) dx$,then find the value of $\int_0^{5T} f(2x) dx$.

Let $f(x)$ be a positive function,$I_1 = \int_{-\frac{1}{2}}^1 2x f(2x(1-2x)) dx$,and $I_2 = \int_{-1}^2 f(x(1-x)) dx$. Then the value of $\frac{I_2}{I_1}$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module