Daily Integral 17 February 2026
Wow, all of today’s integrals are so easy. The medium was only a binomial expansion simplification or using polynomial division. This medium is not worthy of medium. Done the first hard integral ever too.
∫ 0 1 ( x + 1 ) ( x 2 − x + 1 ) d x = ∫ 0 1 x 3 − x 2 + x + x 2 − x + 1 d x = [ 1 4 x 4 + x ] 0 1 = 5 4 \begin{gathered}
\int_0^1 (x+1)(x^2-x+1)\ dx \\
=\int_0^1 x^3-x^2+x+x^2-x+1\ dx \\
=\left[\frac{1}{4}x^4+x\right]_0^1 \\
=\boxed{\frac{5}{4}}
\end{gathered}
∫ 0 1 ( x + 1 ) ( x 2 − x + 1 ) d x = ∫ 0 1 x 3 − x 2 + x + x 2 − x + 1 d x = [ 4 1 x 4 + x ] 0 1 = 4 5
∫ 0 π 2 sin 2 x − sin 4 x d x = ∫ 0 π 2 sin 2 x ( 1 − sin 2 x ) d x = ∫ 0 π 2 sin 2 x cos 2 x d x = 1 4 ∫ 0 π 2 sin 2 2 x d x = 1 8 ∫ 0 π 2 1 − cos 4 x d x = 1 8 [ x − 1 4 sin 4 x ] 0 π 2 = π 16 \begin{gathered}
\int_0^\frac{\pi}{2} \sin^2 x - \sin^4 x\ dx \\
= \int_0^\frac{\pi}{2} \sin^2 x (1 - \sin^2 x)\ dx \\
= \int_0^\frac{\pi}{2} \sin^2 x \cos^2 x\ dx \\
= \frac{1}{4} \int_0^\frac{\pi}{2} \sin^2 2x\ dx \\
= \frac{1}{8} \int_0^\frac{\pi}{2} 1 - \cos 4x\ dx \\
= \frac{1}{8} \left[x - \frac{1}{4}\sin 4x\right]_0^\frac{\pi}{2} \\
= \boxed{\frac{\pi}{16}}
\end{gathered}
∫ 0 2 π sin 2 x − sin 4 x d x = ∫ 0 2 π sin 2 x ( 1 − sin 2 x ) d x = ∫ 0 2 π sin 2 x cos 2 x d x = 4 1 ∫ 0 2 π sin 2 2 x d x = 8 1 ∫ 0 2 π 1 − cos 4 x d x = 8 1 [ x − 4 1 sin 4 x ] 0 2 π = 1 6 π
∫ 0 1 x 4 − 4 x 3 + 6 x 2 − 4 x + 1 x 3 − 3 x 2 + 3 x − 1 x 2 − 2 x + 1 x − 1 d x = ∫ 0 1 ( x − 1 ) 4 ( x − 1 ) 3 ( x − 1 ) 2 ( x − 1 ) d x = ∫ 0 1 ( x − 1 ) 2 d x = ∫ 0 1 x 2 − 2 x + 1 d x = [ 1 3 x 3 − x 2 + x ] 0 1 = 1 3 \begin{gathered}
\int_0^1 \frac{x^4-4x^3+6x^2-4x+1}{\frac{x^3-3x^2+3x-1}{\frac{x^2-2x+1}{x-1}}}\ dx \\
= \int_0^1 \frac{(x-1)^4}{\frac{(x-1)^3}{\frac{(x-1)^2}{(x-1)}}}\ dx \\
= \int_0^1 (x-1)^2\ dx\\
= \int_0^1 x^2-2x+1\ dx\\
= \left[\frac{1}{3}x^3-x^2+x\right]_0^1 \\
= \boxed{\frac{1}{3}}
\end{gathered}
∫ 0 1 x − 1 x 2 − 2 x + 1 x 3 − 3 x 2 + 3 x − 1 x 4 − 4 x 3 + 6 x 2 − 4 x + 1 d x = ∫ 0 1 ( x − 1 ) ( x − 1 ) 2 ( x − 1 ) 3 ( x − 1 ) 4 d x = ∫ 0 1 ( x − 1 ) 2 d x = ∫ 0 1 x 2 − 2 x + 1 d x = [ 3 1 x 3 − x 2 + x ] 0 1 = 3 1
∫ 0 1 ∑ n = 1 ∞ ⌊ 2 n x ⌋ 3 n d x = ∑ n = 1 ∞ ∫ 0 1 ⌊ 2 n x ⌋ 3 n d x = ∑ n = 1 ∞ ∑ i = 1 2 n i − 1 2 n ⋅ 3 n = ∑ n = 1 ∞ 1 2 n ⋅ 3 n ⋅ 2 n ( 2 n − 1 ) 2 = ∑ n = 1 ∞ 2 n − 1 2 ⋅ 3 n = 1 2 ( ∑ n = 1 ∞ ( 2 3 ) n − ∑ n = 1 ∞ 1 3 n ) = 1 2 ( 2 3 1 − 2 3 − 1 3 1 − 1 3 ) = 1 2 ( 2 − 1 2 ) = 3 4 \begin{gathered}
\int_0^1\sum_{n=1}^\infty \frac{\lfloor 2^n x\rfloor}{3^n}\ dx \\
= \sum_{n=1}^\infty \int_0^1 \frac{\lfloor 2^n x\rfloor}{3^n}\ dx \\
= \sum_{n=1}^{\infty} \sum_{i=1}^{2^n} \frac{i - 1}{2^n \cdot 3^n} \\
= \sum_{n=1}^{\infty} \frac{1}{2^n \cdot 3^n} \cdot \frac{2^n(2^n-1)}{2} \\
= \sum_{n=1}^{\infty} \frac{2^n - 1}{2 \cdot 3^n}\\
= \frac{1}{2} (\sum_{n=1}^{\infty} (\frac{2}{3})^n - \sum_{n=1}^{\infty} \frac{1}{3^n}) \\
= \frac{1}{2} (\frac{\frac{2}{3}}{1 - \frac{2}{3}} - \frac{\frac{1}{3}}{1 - \frac{1}{3}}) \\
= \frac{1}{2} (2 - \frac{1}{2}) \\
= \boxed{\frac{3}{4}}
\end{gathered}
∫ 0 1 n = 1 ∑ ∞ 3 n ⌊ 2 n x ⌋ d x = n = 1 ∑ ∞ ∫ 0 1 3 n ⌊ 2 n x ⌋ d x = n = 1 ∑ ∞ i = 1 ∑ 2 n 2 n ⋅ 3 n i − 1 = n = 1 ∑ ∞ 2 n ⋅ 3 n 1 ⋅ 2 2 n ( 2 n − 1 ) = n = 1 ∑ ∞ 2 ⋅ 3 n 2 n − 1 = 2 1 ( n = 1 ∑ ∞ ( 3 2 ) n − n = 1 ∑ ∞ 3 n 1 ) = 2 1 ( 1 − 3 2 3 2 − 1 − 3 1 3 1 ) = 2 1 ( 2 − 2 1 ) = 4 3
