Question

Evaluate the integral as limit of a sum: $\int\limits^3_0 {(e^{x}+6x^{2})} \, dx$

Answer 1

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Answer 2

Answer:

e^3+53

Step-by-step explanation:

We have to integrate the following equation

$\int\limits^3_0 ( {e^x + 6{x^2}}) \, dx$

so for that first integrate the first term wtih respect to x :

$\int\limits^3_0 {e^x} \, dx \\= e^3- e^0\\=e^3-1$ along with the limits 3 as upper limit and 0 as lower limit

and for the another term again integrate with respect to x, we get$\int\limits^3_0 {6(x^2)} \, dx \\=6\times \frac{(x^3)}{3} \\=2\times (x^3)$

and by putting limits we get

$2\times (3^3-0^3)\\=2\times (27-0)\\=54$

and then adding these two equations we get

$\int\limits^3_0 ( {e^x + 6{x^2}}) \, dx$ =$e^3 - 1 + 54\\=e^3+53$

hence the solution is $e^3+53$