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  • Evaluate the definite integrals in Exercises 1 to 20. Integral - 1 to 1 (x + 1)dx

Evaluate the definite integrals in Exercises 1 to 20.

    Q1.    \int_{-1}^{1} (x+1)dx

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Given integral: I = \int_{-1}^{1} (x+1)dx

Consider the integral \int (x+1)dx

\int (x+1)dx = \frac{x^2}{2}+x

So, we have the function of x, f(x) = \frac{x^2}{2}+x

Now, by Second fundamental theorem of calculus, we have

I = f(1)-f(-1)

= \left ( \frac{1}{2}+1\right ) - \left (\frac{1}{2}-1 \right ) = \frac{1}{2}+1-\frac{1}{2}+1 = 2

Posted by

Divya Prakash Singh

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