Evaluate the following integral using the Fundamental Theorem of Calculus. $$int_{1}^{4}(6x^2+7x) dx$$ $$int_{1}^{4}(6x^2+7x) dx = square$$
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The antiderivative of \(6x^2\) is \(2x^3\) and the antiderivative of \(7x\) is \(3.5x^2\). So, the antiderivative of the function is \(2x^3 + 3.5x^2\). Show more…
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