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  • Evaluate the definite integrals in Exercises 1 to 20. Integral 1 to 2 (4x^3 - 5x^2 + 6x + 9)dx

Evaluate the definite integrals in Exercises 1 to 20.

    Q3.    \int_1^2(4x^3-5x^2 + 6x +9)dx

Answers (1)

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Given integral: I = \int_1^2(4x^3-5x^2 + 6x +9)dx

Consider the integral I = \int (4x^3-5x^2 + 6x +9)dx

\int (4x^3-5x^2 + 6x +9)dx = 4\frac{x^4}{4} -5\frac{x^3}{3}+6\frac{x^2}{2}+9x

= x^4 -\frac{5x^3}{3}+3x^2+9x

So, we have the function of x, f(x) = x^4 -\frac{5x^3}{3}+3x^2+9x

Now, by Second fundamental theorem of calculus, we have

I = f(2)-f(1)

=\left \{ 2^4-\frac{5(2)^3}{3}+3(2)^2+9(2)\right \} - \left \{ 1^4-\frac{5(1)^3}{3}+3(1)^2+9(1) \right \}

=\left \{ 16-\frac{40}{3}+12+18\right \} - \left \{ 1-\frac{5}{3}+3+9 \right \}

=\left \{ 46-\frac{40}{3}\right \} - \left \{ 13-\frac{5}{3}\right \}

=\left \{ 33-\frac{35}{3} \right \} = \left \{ \frac{99-35}{3} \right \}

= \frac{64}{3}

Posted by

Divya Prakash Singh

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