JEE Mains · Maths · STD 12 - 7.2 definite integral
Let \(f\left( x \right) = \int\limits_0^x {g\left( t \right)dt} \), where \(g\) is a non zero even function. If \(f(x+5) = g(x)\) , then \(\int\limits_0^x {f\left( t \right)dt} \) equals
- A \(\int\limits_{x + 5}^5 {g\left( t \right)dt} \)
- B \(2\int\limits_{5}^{x - 5} {g\left( t \right)dt} \)
- C \(\int\limits_{5}^{x + 5} {g\left( t \right)dt} \)
- D \(5\int\limits_{x + 5}^5 {g\left( t \right)dt} \)
Answer & Solution
Correct Answer
(A) \(\int\limits_{x + 5}^5 {g\left( t \right)dt} \)
Step-by-step Solution
Detailed explanation
since \(g(x)\) is even with \(f(0)=0\) \(f(x)\) is odd function \(g(x)=f(x+5)\) \(g(-x)=f(-x+5)\) \(g(x)=-f(x-5)\) Replace \(x\) by \(x+5\) \(\Rightarrow f(x)=-g(x+5)\) \(\int_{0}^{x} f(t) d t=-\int_{0}^{x} g(t+5) d t\) \(=-\int_{5}^{x+5} g(t) d t\) \(=\int_{x+5}^{5} g(t) d t\)
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