MCQ
If $f(x) = \int_a^x {{t^3}{e^t}\,dt\,,} $ then $\frac{d}{{dx}}\,f(x) = $
- A${e^x}({x^3} + 3{x^2})$
- ✓${x^3}{e^x}$
- C${a^3}{e^a}$
- DNone of these
✓
Answer
Correct option: B.
${x^3}{e^x}$
b
(b) $f(x) = \int_a^x {{t^3}{e^t}dt = \int_a^0 {{t^3}.{e^t}dt + \int_0^x {{t^3}{e^t}\,\,dt} } } $
(b) $f(x) = \int_a^x {{t^3}{e^t}dt = \int_a^0 {{t^3}.{e^t}dt + \int_0^x {{t^3}{e^t}\,\,dt} } } $
$ \Rightarrow \frac{{df(x)}}{{dx}} = \frac{d}{{dx}}\left( {\int_a^0 {{t^3}.{e^t}dt} } \right) + \frac{d}{{dx}}\left( {\int_0^x {{t^3}.{e^t}\,dt} } \right) = {x^3}{e^x}$.
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