Question

. \(\int\limits_0^3 {{e^{{x^2}}}dx\; = \;\int\limits_0^5 {{e^{{x^2}}}dx + \int\limits_5^3 {{e^{{x^2}}}dx} } } \)

Short answer

The answer is TRUE.

Step-by-step solution

Step 1: The Limit changing property

\(\int\limits_0^3 {{e^{{x^2}}}dx\; = \;\int\limits_0^5 {{e^{{x^2}}}dx + \int\limits_5^3 {{e^{{x^2}}}dx} } } \)

Use the limit changing property :\(\int\limits_a^b {f(x)\;dx\; = \; - \int\limits_b^a {f(x)\;dx} } \)

\(\begin{array}{l}\int\limits_0^3 {{e^{{x^2}}}dx\; = \;\int\limits_0^5 {{e^{{x^2}}}dx - \int\limits_3^5 {{e^{{x^2}}}dx} } } \\\int\limits_0^5 {{e^{{x^2}}}\;dx\; = \;} \int\limits_0^3 {{e^{{x^2}}}dx - \int\limits_3^5 {{e^{{x^2}}}dx} } \end{array}\)

Step 2: The Definite Integral Property

\(\int\limits_a^c {f(x)\;dx\; = \;\int\limits_a^b {f(x)\;dx + \int\limits_b^c {f(x)\;dx} } } \)

Here\(f(x) = {e^{{x^2}}},\;a = 0,b = 3\;and\;c = 5\). So the given statement is true