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Calculus · real student question

Integrate f(x) = x from 0 to 4.

Question

Evaluate

∫04x dx\int_{0}^{4}x\,dx

Step-by-step solution

  1. Recognise which rule applies. The integrand xx is a power x1x^{1}, so the power rule ∫xndx=xn+1n+1+C\int x^{n}dx=\dfrac{x^{n+1}}{n+1}+C with n=1n=1 handles it — no substitution or parts needed.

  2. Write down the antiderivative.

    ∫x dx=x22+C\int x\,dx=\frac{x^{2}}{2}+C

    For a definite integral the constant CC can be dropped: it appears at both limits and cancels in the subtraction.

  3. Apply the Fundamental Theorem of Calculus. Evaluate the antiderivative at the upper limit and subtract its value at the lower limit:

    ∫04x dx=[x22]04=422−022=162−0=8\int_{0}^{4}x\,dx=\left[\frac{x^{2}}{2}\right]_{0}^{4}=\frac{4^{2}}{2}-\frac{0^{2}}{2}=\frac{16}{2}-0=8

  4. Check the result geometrically. On [0,4][0,4] the graph of y=xy=x is a straight line, so the region under it is a right triangle with base 44 and height 44:

    Area=12⋅4⋅4=8\text{Area}=\frac{1}{2}\cdot 4\cdot 4=8

    The geometric area matches the antiderivative computation exactly, which is the strongest available confirmation for a linear integrand.

  5. State the answer.

    ∫04x dx=8\int_{0}^{4}x\,dx=8

Answer

∫04x dx=8\int_{0}^{4}x\,dx = 8

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