Question

If I= $\int\limits_{0}^{2}{{{e}^{{{x}^{4}}}}(x-a)dx}=0$ then a lies in the interval
A) (0, 2)
B) (-1, 0)
C) (2,3)
D) (-2,-1)

Answer 1

Hint:
Separate the integration terms in $\int\limits_{0}^{2}{{{e}^{{{x}^{4}}}}(x-a)dx}=0$ and solve the equation.
Formula used:
$\int\limits_{a}^{b}{}f(x)dx=\left[ F(x) \right]_{a}^{b}=F(b)-F(a)$

Complete step by step solution:
We are given that I= $\int\limits_{0}^{2}{{{e}^{{{x}^{4}}}}(x-a)dx}=0$
$\Rightarrow $ $\int\limits_{0}^{2}{({{e}^{{{x}^{4}}}}\times x-{{e}^{{{x}^{4}}}}\times a)}dx$ = 0
$\Rightarrow $$\int\limits_{0}^{2}{x{{e}^{{{x}^{4}}}}dx-\int\limits_{0}^{2}{a{{e}^{{{x}^{4}}}}dx}}$=0
$\Rightarrow $$\int\limits_{0}^{2}{x{{e}^{{{x}^{4}}}}dx}$=$\int\limits_{0}^{2}{a{{e}^{{{x}^{4}}}}dx}$
These two integration can be same only if a =x
We know that x lies in the interval (0, 2) as given integration is the definite integration lies in the interval of (0,2)

Hence, the interval of ‘a’ also lies between (0, 2).

Additional Information:
In mathematics, an integral assigns numbers to functions in a way that can describe displacement, area, volume, and other concepts that arise by combining infinitesimal data. Integration is one of the two main operations of calculus; its inverse operation, differentiation, is the other. Given a function f of a real variable x and an interval [a, b] of the real line, the definite integral
$\int_{a}^{b}{f}(x)dx$
can be interpreted informally as the signed area of the region in the xy-plane that is bounded by the graph of f, the x-axis and the vertical lines x = a and x = b. The area above the x-axis adds to the total and that below the x-axis subtracts from the total.
$\int\limits_{a}^{b}{}f(x)dx=\left[ F(x) \right]_{a}^{b}=F(b)-F(a)$
Some of integral formulas are
\[\begin{align}
  & \int \text{ }1\text{ }dx\text{ }=\text{ }x\text{ }+\text{ }C \\
 & \int \text{ }a\text{ }dx\text{ }=\text{ }ax+\text{ }C \\
 & \int \text{ }{{x}^{n~}}dx\text{ }=\text{ }\left( \left( {{x}^{n+1}} \right)/\left( n+1 \right) \right)+C\text{ };\text{ }n\ne 1 \\
 & \int \text{ }sin\text{ }x\text{ }dx\text{ }=\text{ }\text{ }cos\text{ }x\text{ }+\text{ }C \\
 & \int \text{ }cos\text{ }x\text{ }dx\text{ }=\text{ }sin\text{ }x\text{ }+\text{ }C \\
 & \int \text{ }se{{c}^{2~}}dx\text{ }=\text{ }tan\text{ }x\text{ }+\text{ }C \\
 & \int \text{ }cs{{c}^{2~}}dx\text{ }=\text{ }-cot\text{ }x\text{ }+\text{ }C \\
 & \int \text{ }sec\text{ }x\text{ }\left( tan\text{ }x \right)\text{ }dx\text{ }=\text{ }sec\text{ }x\text{ }+\text{ }C \\
 & \int \text{ }csc\text{ }x\text{ }\left( \text{ }cot\text{ }x \right)\text{ }dx\text{ }=\text{ }\text{ }csc\text{ }x\text{ }+\text{ }C \\
 & \int \text{ }\left( 1/x \right)\text{ }dx\text{ }=\text{ }ln\text{ }\left| x \right|\text{ }+\text{ }C \\
 & \int \text{ }{{e}^{x~}}dx\text{ }=~{{e}^{x}}+\text{ }C \\
 & \int \text{ }{{a}^{x~}}dx\text{ }=\text{ }\left( {{a}^{x}}/ln\text{ }a \right)\text{ }+\text{ }C\text{ };\text{ }a>0,~~a\ne 1 \\
\end{align}\]

Note:
The knowledge about integration (definite integral and indefinite integral) is important for students to answer such questions.