If $f\,\left( x \right)$ is continuous for all real values of $x$, then $\mathop \sum \limits_{r = 1}^{10} \mathop {\,\,\smallint }\nolimits_0^1 \left( {r - 1 + x} \right)\,\,dx$ is equal to

$\mathop \smallint \nolimits_0^{10} \,\,f\left( x \right)\,\,dx$

$\mathop \smallint \nolimits_0^1 \,\,f\left( x \right)\,\,dx$

$10\,\,\mathop \smallint \nolimits_0^1 \,\,f\left( x \right)\,\,dx$

$9\mathop {\,\,\smallint }\nolimits_0^1 \,\,f\left( x \right)\,\,dx$

JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

1 Easy

1 of 1

Analyzing +4 | -1

f\,\left( x \right)

is continuous for all real values of

x

, then

\sum \limits_{r = 1}^{10} \mathop {\,\,\int }\nolimits_0^1 \left( {r - 1 + x} \right)\,\,dx

is equal to

View answer and explanation

Correct option: A

\int \nolimits_0^{10} \,\,f\left( x \right)\,\,dx

Option A (

\int\nolimits_0^{10} \,\,f\left( x \right)\,\,dx

) is correct because the sum of the integrals simplifies to a single integral from 0 to 10 due to the continuity of f(x).