Someone computes ∫_-1^1(dx)/(x^2)=Big[-1xBig]_-1^1=-2. What is the error?

1/x^2 is neither continuous nor bounded at 0, so the theorem does not apply

The antiderivative -1/x is wrong

No error: the integral equals -2

Exercise · ZW-FUNDAMENTAL-THEOREM-OF-CALCULUS-020 · ★★★ Advanced

Someone computes \(\displaystyle\int_{-1}^{1}\frac{dx}{x^2}=\Big[-\frac1x\Big]_{-1}^{1}=-2\). What is the error?

Choose one option.

Show answer

D. \(1/x^2\) is neither continuous nor bounded at \(0\), so the theorem does not apply

Explanation

The integrand is positive, so no integral could be negative. The theorem requires \(f\) continuous on all of \([-1,1]\); here \(1/x^2\) is unbounded near \(0\) and the Riemann integral does not exist.

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Review the concept: Fundamental theorem of calculus

It links derivatives and integrals: a definite integral equals an antiderivative evaluated at the endpoints, and differentiating an integral with variable upper limit returns the integrand.

\[\begin{gathered}\int_a^b f(x)\,dx=F(b)-F(a)\\ \frac{d}{dx}\int_a^x f(t)\,dt=f(x)\end{gathered}\]

Someone computes \(\displaystyle\int_{-1}^{1}\frac{dx}{x^2}=\Big[-\frac1x\Big]_{-1}^{1}=-2\). What is the error?

Explanation

Review the concept: Fundamental theorem of calculus

More practice problems: fundamental theorem of calculus